Statement:-
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) then there exists at least one c in (a,b) such that f’(c)/g'(c) = f(b)-f(a)/g(b)-g(a).
Proof:-
Let us define a function
F(x) = f(x){g(b)-g(a)} - g(x){f(b)-f(a)}
This function is continious in [a,b] and differentiable in (a,b), since f(x) and x are continious in [a,b] and differentiable in (a,b).
Here F(a) = f(a){g(b)-g(a)} - g(a){f(b)-f(a)}
= f(a)g(b)-g(a)f(b)
F(b) = f(b){g(b)-g(a)} - g(b){f(b)-f(a)} = F(a)
The by Rolle's Theorem there exist at least one point c in (a,b) such that F'(c) = 0.
f'(c){g(b)-g(a)} = g'(c){f(b)-f(a)}
f’(c)/g'(c) = f(b)-f(a)/g(b)-g(a) g'(x) not equal to 0 in (a,b) and g(b) not equal to g(a)
Monday, June 21, 2010
Lagrange's Theorem
Statement:-
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) then there exists at least one c in (a,b) such that f’(c) = f(b)-f(a)/b-a.
Proof:-
Let us define a function
F(x) = f(x)(b-a)-[f(b)-f(a)]x........(i)
which is continious in [a,b] and differentiable in (a,b), since f(x) and x are continious in [a,b] and differentiable in (a,b).
Here F(a) = f(a)(b-a)-[f(b)-f(a)]a
= bf(a)-af(b)-af(b)+af(a)=bf(a) - af(b)
F(b) = f(b)(b-a)-[f(b)-f(a)]b
= bf(b)-af(b)-bf(b)+bf(a)
=bf(a)-af(b) = F(a)
The by Rolle's Theorem there exist at least one point c in (a,b) such that F'(c) = 0.
But F'(c) = f'(c)(b-a)-[f(b)-f(a)].
f'(c)(b-a)-[f(b)-f(a)] = 0
f'(c) = f(b)-f(a)/b-a
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) then there exists at least one c in (a,b) such that f’(c) = f(b)-f(a)/b-a.
Proof:-
Let us define a function
F(x) = f(x)(b-a)-[f(b)-f(a)]x........(i)
which is continious in [a,b] and differentiable in (a,b), since f(x) and x are continious in [a,b] and differentiable in (a,b).
Here F(a) = f(a)(b-a)-[f(b)-f(a)]a
= bf(a)-af(b)-af(b)+af(a)=bf(a) - af(b)
F(b) = f(b)(b-a)-[f(b)-f(a)]b
= bf(b)-af(b)-bf(b)+bf(a)
=bf(a)-af(b) = F(a)
The by Rolle's Theorem there exist at least one point c in (a,b) such that F'(c) = 0.
But F'(c) = f'(c)(b-a)-[f(b)-f(a)].
f'(c)(b-a)-[f(b)-f(a)] = 0
f'(c) = f(b)-f(a)/b-a
Rolle’s Theorem
Statement:-
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) and f(a) = f(b), there exists at least one c in (a,b) such that f’(c) = 0.
Proof:-
Since f(x) is continuous in closed interval [a,b], there exist c,d in [a,b] such that
f(c) = M(the maximum value of f(x))
f(d) = m the minimum value of f(x)
CASE I
If M = m, then f(x) is constant in [a,b] and in this case f’(x) = 0 for all x in (a,b). That is theorem is true in this case.
CASE II
If M = m, then at least one of them is different from f(a) and f(b). Let M = f(a). We shall show that f’(c) = 0 with c in (a,b)
At x = c, with h>0
RHD = lim h 0 f(c+h)-f(c)/h
= lim h 0 f(c+h)-M/h = -ve/+ve <= 0 ……….i
LHD = lim h 0 f(c-h)-f(c)/-h
= lim h 0 f(c-h)-M/-h = -ve/-ve >= 0 ……….ii
Since f’(x) exist in (a,b), I and ii must be equal and this will be possible only when both i and ii are equal to zero.
Hence f’(c) = 0 with c in (a,b).
CASE III
If m is different from f(a) and f(b) we can similarly show that f’(d) = 0 with d in (a,b). This completes the proof.
If f(x) is a continuous in a closed interval [a,b], differentiable in (a,b) and f(a) = f(b), there exists at least one c in (a,b) such that f’(c) = 0.
Proof:-
Since f(x) is continuous in closed interval [a,b], there exist c,d in [a,b] such that
f(c) = M(the maximum value of f(x))
f(d) = m the minimum value of f(x)
CASE I
If M = m, then f(x) is constant in [a,b] and in this case f’(x) = 0 for all x in (a,b). That is theorem is true in this case.
CASE II
If M = m, then at least one of them is different from f(a) and f(b). Let M = f(a). We shall show that f’(c) = 0 with c in (a,b)
At x = c, with h>0
RHD = lim h 0 f(c+h)-f(c)/h
= lim h 0 f(c+h)-M/h = -ve/+ve <= 0 ……….i
LHD = lim h 0 f(c-h)-f(c)/-h
= lim h 0 f(c-h)-M/-h = -ve/-ve >= 0 ……….ii
Since f’(x) exist in (a,b), I and ii must be equal and this will be possible only when both i and ii are equal to zero.
Hence f’(c) = 0 with c in (a,b).
CASE III
If m is different from f(a) and f(b) we can similarly show that f’(d) = 0 with d in (a,b). This completes the proof.
Sunday, June 13, 2010
Self Induction
When there is change in current through a solonoid then there will be induced emf in opposite direction of cause which produces it. This process is called as self induction
Now flux linked with the coil of solonoid
φ α I
φ = LI----------1 where L is called inductance of coil
Now we have
e = -dφ/dt
e = -d(LI)/dt
= -LdI/dt--------2
If dI/dT = 1amp/sec
Then e=L
Hence inductance of a solonoid can be defined as the back emf induced when the rate of change of current is 1 unit
Self inductance of a solonoid
Let a solonoid of length L having total no, of turns of coil N
Let Current I passes through it
then B = UonI
= UoN/l *I
φ = BAN
= UoAN^2I/L
we have
L = φ/I = UoAN^2/L
Now flux linked with the coil of solonoid
φ α I
φ = LI----------1 where L is called inductance of coil
Now we have
e = -dφ/dt
e = -d(LI)/dt
= -LdI/dt--------2
If dI/dT = 1amp/sec
Then e=L
Hence inductance of a solonoid can be defined as the back emf induced when the rate of change of current is 1 unit
Self inductance of a solonoid
Let a solonoid of length L having total no, of turns of coil N
Let Current I passes through it
then B = UonI
= UoN/l *I
φ = BAN
= UoAN^2I/L
we have
L = φ/I = UoAN^2/L
Brewsters Law
We can get polarized light from reflection according to Brewsters law which states that when a light incident on a glass slab at certain angle then, we can get the polarized light in reflected part polarized in direction perpendicular to plane of incidence. This particular angle is called polarizing angle and in this case the sum of angle of refraction and angle of polarization will be 90 i.e angle between reflected and refracted rays will be 90
p+r = 90
i.e r = 90
Now according to snell's lae
refractive index = sini/sinr
= sinp/sin(90-p)
= sinp/cosp
=tanp
Therefore refractive index of material is equal to tangent of polarizing angle.
Malus law
The intensity of the light coming out from analyzer will be directly proportional to the square of cosine of angle between polarizer and analyzer.
Now, if a be the amplitude of the light used and (theta) be the angle between analyzer and polarizer then acos(theta) will be the component of a along analyzer and asin(theta) will be that along perpendicular direction. Therefore intensity of light which pass through analyzer is
I = (acos(theta))^2
since a is constant
I α cos^2(theta)
Calcite crystal
It is a cuboid made up of six 119m having angle 109 and 71. There will be two corners in which all parallepgram will meet at obtuse angle are called blunt corners. The axis of crystal. The light incident on optical axis will pass without splitting. In actual practice light incident parallel to potical axis will pass without splitting. SO that optical axis is not just a line, it is a direction
Principal section
Principal section is a plane which contains optical axis and perpendicular to opposite faces
Quarter wave plate
A special type of crystal which introduces a path difference of λ/4(i.e quarter of wavelength) between two emergent light is called quarter wave plate
optical path for o-ray = Uot
optical path for e-ray = Uet
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/4
Half wave plate
A special type of crystal which introduces path diference of λ/2(i.e) half of wave length) between two emergent light is called half wave plate
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/2
p+r = 90
i.e r = 90
Now according to snell's lae
refractive index = sini/sinr
= sinp/sin(90-p)
= sinp/cosp
=tanp
Therefore refractive index of material is equal to tangent of polarizing angle.
Malus law
The intensity of the light coming out from analyzer will be directly proportional to the square of cosine of angle between polarizer and analyzer.
Now, if a be the amplitude of the light used and (theta) be the angle between analyzer and polarizer then acos(theta) will be the component of a along analyzer and asin(theta) will be that along perpendicular direction. Therefore intensity of light which pass through analyzer is
I = (acos(theta))^2
since a is constant
I α cos^2(theta)
Calcite crystal
It is a cuboid made up of six 119m having angle 109 and 71. There will be two corners in which all parallepgram will meet at obtuse angle are called blunt corners. The axis of crystal. The light incident on optical axis will pass without splitting. In actual practice light incident parallel to potical axis will pass without splitting. SO that optical axis is not just a line, it is a direction
Principal section
Principal section is a plane which contains optical axis and perpendicular to opposite faces
Quarter wave plate
A special type of crystal which introduces a path difference of λ/4(i.e quarter of wavelength) between two emergent light is called quarter wave plate
optical path for o-ray = Uot
optical path for e-ray = Uet
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/4
Half wave plate
A special type of crystal which introduces path diference of λ/2(i.e) half of wave length) between two emergent light is called half wave plate
In case of clacite crystal Ue
In case of quarts crystal Ue>Uo
Uet - Uot = λ/2
Dopplar's effect
When source of sound and observer in motion either or both (or relative motion) then their will be apparent change in frequency. This phenomenon is known as Doppler's effect.
CaseI : Sound in motion observer in rest
Let us consider a source of sound which generates sound of frequency f wavelength λ and velocity V be moving toward a stationary observer O with velocity Us. Then observer feels apparent change in wave length as
λ = (V-Us)/f--------------1
Now relative motion of wave with respect to observer will be V - 0 = V
So the apparent frequency observered by the observer will be
f1 = v /λ = V/(V-Us) * f-----------2
Here V-Uo < U so clearly f1 > f SO loud sound is heared when the source moves towards stationary observer.
CASEII: When source is moving away the observer then Us will be negative and hence apparent freq.
f1 = V/(V+Us) * f--------------3
CASEIII : WHen observer in motion and source in rest
Let an observer O be in motion and source is in rest. Let the observer wave toward the source with velocity Vo, let the velocity of sound wave be V. Wave length λ and frequency f
In this case the wave length of sound remains unchanged and be given as V/P - λ
Here since the observer is moving towards source relative velocity of wave with respect to observer is V+Vo
Therefore the apparent change in frequency will be
f1 = relativevelocity/wave length = V+Vo\λ
.-.f1 = (V+Uo)/V * f------------4
Since V+Vo is greater than V the pitch of sound increases and sound will be heard louder
CASE IV:
If the observer is moving away from source relative velocity will be V-Vo apparent freq.
f1 = (V-Vo)/V * f --------------5
CaseI : Sound in motion observer in rest
Let us consider a source of sound which generates sound of frequency f wavelength λ and velocity V be moving toward a stationary observer O with velocity Us. Then observer feels apparent change in wave length as
λ = (V-Us)/f--------------1
Now relative motion of wave with respect to observer will be V - 0 = V
So the apparent frequency observered by the observer will be
f1 = v /λ = V/(V-Us) * f-----------2
Here V-Uo < U so clearly f1 > f SO loud sound is heared when the source moves towards stationary observer.
CASEII: When source is moving away the observer then Us will be negative and hence apparent freq.
f1 = V/(V+Us) * f--------------3
CASEIII : WHen observer in motion and source in rest
Let an observer O be in motion and source is in rest. Let the observer wave toward the source with velocity Vo, let the velocity of sound wave be V. Wave length λ and frequency f
In this case the wave length of sound remains unchanged and be given as V/P - λ
Here since the observer is moving towards source relative velocity of wave with respect to observer is V+Vo
Therefore the apparent change in frequency will be
f1 = relativevelocity/wave length = V+Vo\λ
.-.f1 = (V+Uo)/V * f------------4
Since V+Vo is greater than V the pitch of sound increases and sound will be heard louder
CASE IV:
If the observer is moving away from source relative velocity will be V-Vo apparent freq.
f1 = (V-Vo)/V * f --------------5
Faradays Lenz
Faraday's Law of Electromagnetic induction states that
1) When there is change of flux linked iwth a conductor then emf will be induced.
2) Induced emf will continue till the change in flux continues.
3) The induced emf is directly proportional to change of flux
Lenz law
It states that the induced emf is in the direction so as to oppose the cause producing it
Lenz law is in accordance with conservation of energu. If we are taking a magneti with north pole pointing towards the coil, then the magnetic field of coil due to induced emf will be in near end of magnet which repel magnet. So some work must be done against this force of repulsion which will be converted into electrical energy.
1) When there is change of flux linked iwth a conductor then emf will be induced.
2) Induced emf will continue till the change in flux continues.
3) The induced emf is directly proportional to change of flux
Lenz law
It states that the induced emf is in the direction so as to oppose the cause producing it
Lenz law is in accordance with conservation of energu. If we are taking a magneti with north pole pointing towards the coil, then the magnetic field of coil due to induced emf will be in near end of magnet which repel magnet. So some work must be done against this force of repulsion which will be converted into electrical energy.
Ampere's Circutial law:
It states that the line integral of magnetic field over a closed surface is equal to Uo times the total current enclosed by the surface.
i.e {B.dl = UoI
If we consider a straight conductor I, then it will produce a circular magnetic field. Now magnetic field at a point at a distance r from conductor can be give as
B = UoI/2πr
Now {B.dl = {Bdlcosϴ
={Bdl
= {UoI/2πr dl
=UoI/2πr 0{2πr dl
{B.dl = UoI
Applicatiojn of Ampere's law
1) Magnetic field due to straight conductor
Let us consider a straight conductor carrying current I. Consider a point P at distance R from conductor where magnetic field is to be determined. Since this is straight conductor magnetic lines of force are circular. Let us draw a circle of radius r passing through P.
then from Ampere's Law,
{B.dl = UoI
{Bdlcosϴ = UoI
B = UoI/2πr
2) Magnetic field due to solenoid
Let us consider a solonoid having number of turns per unit length n carry current I through it. It produces magnetic field along its axis Now to find magnetic field draw a rectangle of length PQ = l
By ampere's law
{B.dl = UoI
{B.dl = P{Q B.dl + R{Q B.dl + S{R B.dl + P{S B.dl
{B.dl = B P{Q dl =Bl----------1
Now
{B.dl = UoI
{B.dl = UonlI--------2
From I and 2
B = UonI
i.e {B.dl = UoI
If we consider a straight conductor I, then it will produce a circular magnetic field. Now magnetic field at a point at a distance r from conductor can be give as
B = UoI/2πr
Now {B.dl = {Bdlcosϴ
={Bdl
= {UoI/2πr dl
=UoI/2πr 0{2πr dl
{B.dl = UoI
Applicatiojn of Ampere's law
1) Magnetic field due to straight conductor
Let us consider a straight conductor carrying current I. Consider a point P at distance R from conductor where magnetic field is to be determined. Since this is straight conductor magnetic lines of force are circular. Let us draw a circle of radius r passing through P.
then from Ampere's Law,
{B.dl = UoI
{Bdlcosϴ = UoI
B = UoI/2πr
2) Magnetic field due to solenoid
Let us consider a solonoid having number of turns per unit length n carry current I through it. It produces magnetic field along its axis Now to find magnetic field draw a rectangle of length PQ = l
By ampere's law
{B.dl = UoI
{B.dl = P{Q B.dl + R{Q B.dl + S{R B.dl + P{S B.dl
{B.dl = B P{Q dl =Bl----------1
Now
{B.dl = UoI
{B.dl = UonlI--------2
From I and 2
B = UonI
Biot-Savart Law
It states that if a conductor carry current I having length l produces magnetic field that is directly proportional to I sinϴ dl and inversly proportional to square of r.
Let us consider a wire of length lcarries conductor I through it. Let us consider a point P at a distance r from an elementary length dl And ϴ be the angle made by r with dl. Then there will be magnetic field produced at P.
Now Biot and Savart kaw states that:
Magnetic field produced at p due to elementry length dl is
1) directly proportional to I
dB α I
2)directly proportional to sine of ϴ
dB α sinϴ
3)directly proportional to dl
dB α dl
4)inversly proportional to the square of r
dB α 1/r^2
Therefor
dB α Idlsinϴ/r^2
dB = KIdlsinϴ/r^2
dB = Uo/4π * Idlsinϴ/r^2
Application of Biot-Savart's Law
1) Magnetic field at centre of circular coil
Let us consider a circular coil of radius r carrying current in the direction as shown in figure. To calculate the magnetic field at centre of this coil, consider an elementary length dl which makes an angle of ϴ with radius vector r.
Now according to Biot-savart law the magnetic field at the centre of coil can be given as,
dB = Uo/4π * Idlsinϴ/r^2
Now, total magnetic field at centre due to whole coil can be given as,
B = 0[2πr U0/4π Idlsinϴ/r^2
B = U0Isinϴ/4πr^2 0[2πr dl
B = UoI/4πr^2 [l] 0-2πr
B = UoI/2r
if there are n coils
B = U0nI/2r
2)Magnetic field on axis of coil
Let us consider a circular coil of radius a which carry current I through it. Let its plane be perpendicular to the plane of paper such that its axis lied on plane of paper. Let us consider a point p at a distance x from centre of coil on its axis. Let us consider an elementary length dl on coil and r be the distance between p and dl. Let β be the angle between dl and r.
Now, magnetic field at P due to this elementary length dl is
dB = Uo/4π * Idlsinϴ/r^2
Here we can take
β =π/2
dB =Uo/4π * Idl/r^2--------------1
Now direction of dB is perpendicular to both dl and r. Let it be along PR which can be resloved into tow componentd dBcosϴ along perpendicular to axis and dBsinϴ along axix. Since coil is symmetric about axis. These perpendicular components, i.e dBcosϴ will cancel each other and therefore their contribution is zero. SO the magnetic field is due to the component along x-axis.
The magnetic field due to whole coil at point P can be given as:
B = [dBsinϴ
B = [Uo/4π * Idlsinϴ/r^2
B = Uo/4π * Isinϴ/r^2 0[2πr dl
B = Uo/4πr^2 Isinϴ*2πa
B = UoIa^2/2r^3
B = UoIa^2/2(a^2+x^2)^3/2
Case I
when P lies at the centre of coil x =0
B = UoIa^2/2a^3
B = UoI/2a
Case II
when P lies at very far disrance a<<<<
3) Magnetic field due to long cunductor.
Let us consider a long conductor carrying current I. Consider a point P at a distance a from centre of conductor where magnetic field is to be found.
Let us consider on elementary length dl,AB. Let P be at a distance r from A and AB subtend an angle dα at P. Again let r makes an angle α with conductor. Now draw a perpendicular AC from A to BP. The we can have
Now in /_\ ACP
sindα = AC/AP
or dα = AC/r
AC = rdα-----------2
Again in /_\ ABC,
sinα = AC/AB = Ac/d1
AC = dlsinα
From 2 and 3
dlsinα = rdα
r = a/sinα
dB = Uo/4π * I/a * sinαdα
Now magnetic field at point P due to whole conductor can be given as
B = α1{α2 Uo/4π * I/a * sinαdα
If we consider the conductor as infinitely long than we can have
B = UoI/4a 0{π sinαdα-----------7
B = UoI/4πa *2
B = UoI/2πa
Let us consider a wire of length lcarries conductor I through it. Let us consider a point P at a distance r from an elementary length dl And ϴ be the angle made by r with dl. Then there will be magnetic field produced at P.
Now Biot and Savart kaw states that:
Magnetic field produced at p due to elementry length dl is
1) directly proportional to I
dB α I
2)directly proportional to sine of ϴ
dB α sinϴ
3)directly proportional to dl
dB α dl
4)inversly proportional to the square of r
dB α 1/r^2
Therefor
dB α Idlsinϴ/r^2
dB = KIdlsinϴ/r^2
dB = Uo/4π * Idlsinϴ/r^2
Application of Biot-Savart's Law
1) Magnetic field at centre of circular coil
Let us consider a circular coil of radius r carrying current in the direction as shown in figure. To calculate the magnetic field at centre of this coil, consider an elementary length dl which makes an angle of ϴ with radius vector r.
Now according to Biot-savart law the magnetic field at the centre of coil can be given as,
dB = Uo/4π * Idlsinϴ/r^2
Now, total magnetic field at centre due to whole coil can be given as,
B = 0[2πr U0/4π Idlsinϴ/r^2
B = U0Isinϴ/4πr^2 0[2πr dl
B = UoI/4πr^2 [l] 0-2πr
B = UoI/2r
if there are n coils
B = U0nI/2r
2)Magnetic field on axis of coil
Let us consider a circular coil of radius a which carry current I through it. Let its plane be perpendicular to the plane of paper such that its axis lied on plane of paper. Let us consider a point p at a distance x from centre of coil on its axis. Let us consider an elementary length dl on coil and r be the distance between p and dl. Let β be the angle between dl and r.
Now, magnetic field at P due to this elementary length dl is
dB = Uo/4π * Idlsinϴ/r^2
Here we can take
β =π/2
dB =Uo/4π * Idl/r^2--------------1
Now direction of dB is perpendicular to both dl and r. Let it be along PR which can be resloved into tow componentd dBcosϴ along perpendicular to axis and dBsinϴ along axix. Since coil is symmetric about axis. These perpendicular components, i.e dBcosϴ will cancel each other and therefore their contribution is zero. SO the magnetic field is due to the component along x-axis.
The magnetic field due to whole coil at point P can be given as:
B = [dBsinϴ
B = [Uo/4π * Idlsinϴ/r^2
B = Uo/4π * Isinϴ/r^2 0[2πr dl
B = Uo/4πr^2 Isinϴ*2πa
B = UoIa^2/2r^3
B = UoIa^2/2(a^2+x^2)^3/2
Case I
when P lies at the centre of coil x =0
B = UoIa^2/2a^3
B = UoI/2a
Case II
when P lies at very far disrance a<<<<
3) Magnetic field due to long cunductor.
Let us consider a long conductor carrying current I. Consider a point P at a distance a from centre of conductor where magnetic field is to be found.
Let us consider on elementary length dl,AB. Let P be at a distance r from A and AB subtend an angle dα at P. Again let r makes an angle α with conductor. Now draw a perpendicular AC from A to BP. The we can have
Now in /_\ ACP
sindα = AC/AP
or dα = AC/r
AC = rdα-----------2
Again in /_\ ABC,
sinα = AC/AB = Ac/d1
AC = dlsinα
From 2 and 3
dlsinα = rdα
r = a/sinα
dB = Uo/4π * I/a * sinαdα
Now magnetic field at point P due to whole conductor can be given as
B = α1{α2 Uo/4π * I/a * sinαdα
If we consider the conductor as infinitely long than we can have
B = UoI/4a 0{π sinαdα-----------7
B = UoI/4πa *2
B = UoI/2πa
Electric Dipole
System of two charges equal in magnitude but opposite in nature separated by certain distance is called an electric dipole.
Dipole moment is given as
P = qL, directed from -ve charge to +ve charge.
Electric potential due to electric dipole
Let us consider an electric dipole Ab of length L formed by charges +q and -q. Let p be a point at a distance r from centre of dipole where electric potential is to be found.
We know
V = V1+V2
V1 = 1/4πEo(q/PB)
V2 = 1/4πEo(q/AP)
if we consider dipole of very short length we can have AP = PN and PM = PB
also we know
PM = r-l/2cos(theta)
PN = r+l/2cos(theta)
Now
V = q/4πEo(1/r-l/2cos(theta) - 1/r+l/2cos(theta))
V = q/4πEo(lcos(theta)/r^2-l^2/4cos^2(theta))
since l^2/4cos^2(theta <<< r^2 we can have
V = pcos(theta)/4πEor^2
Dipole moment is given as
P = qL, directed from -ve charge to +ve charge.
Electric potential due to electric dipole
Let us consider an electric dipole Ab of length L formed by charges +q and -q. Let p be a point at a distance r from centre of dipole where electric potential is to be found.
We know
V = V1+V2
V1 = 1/4πEo(q/PB)
V2 = 1/4πEo(q/AP)
if we consider dipole of very short length we can have AP = PN and PM = PB
also we know
PM = r-l/2cos(theta)
PN = r+l/2cos(theta)
Now
V = q/4πEo(1/r-l/2cos(theta) - 1/r+l/2cos(theta))
V = q/4πEo(lcos(theta)/r^2-l^2/4cos^2(theta))
since l^2/4cos^2(theta <<< r^2 we can have
V = pcos(theta)/4πEor^2
Gauss Law:
The total lines of force(electric flux) passing through a surface is eqyal to 1/Eo times the total charge enclosed by that surface.
φ = 1/Eo * total charge
Application og gauss law
1) To find electric intnensity due to charged sphere
Let us consdier a charged sphere of radius E and p be a point at distance r from the centre of sphere
CASE I when P lies outside the sphere
To find the intensity at { drae a concentric sphere of radius r such that it will enclose charge on given sphere. Let it be q then according to gauss law,
Elelctrc flux, φ = 1/Eo * q
E *A = 1/Eo *q
E = 1/4πr^2 * q/Eo
E = 1/4πE0 *q/r^2
CASE II P lies on surface of sphere
IN this r = R
E = 1/4πE0 *q/R^2
CASE III P lies inside the sphere
In this case the total charge enclosed by a gaussian surface draws through P(sphere) will be zero.
φ = 1/Eo * q = 0
E *A =0
E = 0
II) Intensity due to a line charge
Let us consider a infinitely long conductor which contains the charge as linear charge density λ. Let p be a point at a distance r from the conductor where electric intensity is to be found. To find this let us draw a cylinder through P co-axial with the conductor of length L.
Then. from Gauss law, flux through this cylinder,
φ = 1/Eo * total charge
φ = 1/Eo * λ
E * A = 1/Eo λ L
E * 2πrl = 1/Eo λ l
E = λ/2πEor
III) Intensity due to charged plane sheet
Let us consider a plane charged sheet having surface charge density E. Let P be a point at distance r from plane sheet where intensity is to be found. To find the intensity, let us draw a cylinder enclosing P, perpendicular to plane sheet having surface area A. The form gauss law
φ = 1/Eo * total charge
φ = 1/Eo * 6A
E *A = 1/Eo * 6A
E = 6/Eo
φ = 1/Eo * total charge
Application og gauss law
1) To find electric intnensity due to charged sphere
Let us consdier a charged sphere of radius E and p be a point at distance r from the centre of sphere
CASE I when P lies outside the sphere
To find the intensity at { drae a concentric sphere of radius r such that it will enclose charge on given sphere. Let it be q then according to gauss law,
Elelctrc flux, φ = 1/Eo * q
E *A = 1/Eo *q
E = 1/4πr^2 * q/Eo
E = 1/4πE0 *q/r^2
CASE II P lies on surface of sphere
IN this r = R
E = 1/4πE0 *q/R^2
CASE III P lies inside the sphere
In this case the total charge enclosed by a gaussian surface draws through P(sphere) will be zero.
φ = 1/Eo * q = 0
E *A =0
E = 0
II) Intensity due to a line charge
Let us consider a infinitely long conductor which contains the charge as linear charge density λ. Let p be a point at a distance r from the conductor where electric intensity is to be found. To find this let us draw a cylinder through P co-axial with the conductor of length L.
Then. from Gauss law, flux through this cylinder,
φ = 1/Eo * total charge
φ = 1/Eo * λ
E * A = 1/Eo λ L
E * 2πrl = 1/Eo λ l
E = λ/2πEor
III) Intensity due to charged plane sheet
Let us consider a plane charged sheet having surface charge density E. Let P be a point at distance r from plane sheet where intensity is to be found. To find the intensity, let us draw a cylinder enclosing P, perpendicular to plane sheet having surface area A. The form gauss law
φ = 1/Eo * total charge
φ = 1/Eo * 6A
E *A = 1/Eo * 6A
E = 6/Eo
Ultrasound
The sound wave which has the frequency greater then 20khz i.e beyond the audible region is called the untrasound. The sound wave which has frequency less than 20hz are known as infrasonic sound wave or infrasound.
Production of ultrasound
1) Piezo electric method
When a force is applied into a pair faces of quartz crystal then an electric potential will be setup into the opposite pair faces. Conversly, if an electric potential be applied on a pair of quartz crystal then there will be tension on the opposite faces. This effect in quartz crystal is known as piezo electric effect.
Now When an alternating potential be applied on a pair forces of a quartz crystal then opposite force will be in longitidunal vibration when freq. of applied potential becomes equal to natural frequency of quartz crystal then resonance will occur. In that case crystal will vibrate with maximum amplitude and a sound wave will be produced with ultrasound properties.
2) Magnetic striction method
In this method a ferromagnetic material is wounded closely by a coil and alternating current is passed through it. Then we may experience a longitudinal vibration in the ferromagnetic substance. When the frequency of the alternating current becomes equal to the natural frequency of ferromagnetc substance then resonance will occur and the material will vibrate with maximum amplitude. And a sound wave is generated with unltasound Properties.
Use of ultrasound
1) To find the depth of Seas
2) To locate distance object
3) To kill unwanted tissue(blood less surgery)
Characterstics of Sound
1) Pitch: Pitch is a characterstics of sound which directly related with frequency. Sound wave having high frequency is called high pitched sound and vice versa.
2) Intensity: Sound energy flowing per unit area per unit time is called intensity
3) Threshold frequency of hearing: The minimum value of sound intensity that can be heared by human being in general
4) Loudness: It is the magnitue of sensation of hearing and it is directly proportional to the logarithm of intensity of sound
5) Quality: Quality of sound depends on no. of harmonics presents in sound. HIgher no. of harmonic reveals high quality and vice versa.
Production of ultrasound
1) Piezo electric method
When a force is applied into a pair faces of quartz crystal then an electric potential will be setup into the opposite pair faces. Conversly, if an electric potential be applied on a pair of quartz crystal then there will be tension on the opposite faces. This effect in quartz crystal is known as piezo electric effect.
Now When an alternating potential be applied on a pair forces of a quartz crystal then opposite force will be in longitidunal vibration when freq. of applied potential becomes equal to natural frequency of quartz crystal then resonance will occur. In that case crystal will vibrate with maximum amplitude and a sound wave will be produced with ultrasound properties.
2) Magnetic striction method
In this method a ferromagnetic material is wounded closely by a coil and alternating current is passed through it. Then we may experience a longitudinal vibration in the ferromagnetic substance. When the frequency of the alternating current becomes equal to the natural frequency of ferromagnetc substance then resonance will occur and the material will vibrate with maximum amplitude. And a sound wave is generated with unltasound Properties.
Use of ultrasound
1) To find the depth of Seas
2) To locate distance object
3) To kill unwanted tissue(blood less surgery)
Characterstics of Sound
1) Pitch: Pitch is a characterstics of sound which directly related with frequency. Sound wave having high frequency is called high pitched sound and vice versa.
2) Intensity: Sound energy flowing per unit area per unit time is called intensity
3) Threshold frequency of hearing: The minimum value of sound intensity that can be heared by human being in general
4) Loudness: It is the magnitue of sensation of hearing and it is directly proportional to the logarithm of intensity of sound
5) Quality: Quality of sound depends on no. of harmonics presents in sound. HIgher no. of harmonic reveals high quality and vice versa.
Beat and Beat frequency
When two sound of nearly equal frequency are sounded together then periodic rise and fall in sound intensity is heared. This phenomeno is called beat. For this phenomenon the difference between frequencies of two sound wave must not exceed 10hz
Beat frewuency
Let us consider two sound waves with frequency f1 and f2 such that f1-f2 <10 hz are sounded together then at the starting points can be given as
y1 = asinw1t
y2 = asinw2t
By the principal of superpostion the resultant wave will be,
y = y1+y2
y = asinw1t + asinw2t
y = A sin(pi)(f1+f2) * t
This is a periodic function with amplitude
A = 2acos(pi)(f1-f2) *t
CASE I Loud sound
Loud sound will be heard when value of A becomes maximum i.e when
cos(pi)(f1-f2)*t = 1
cos(pi)(f1-f2)*t = cpsn(pi)
f = n/(f1-f2)
loud sound will be heared at f = 0, 1/(f1-f2),2/(f1-f2)
Beat frequency = (f1-f2) hz
CASEII Soft Sound
Soft sound will be heared when amplitude is minimum i.e when
cos(pi)(f1-f2)*t = cos((2n+1)/2)*pi
f = 1/(f1-f2)
Beat frequency = (f1-f2)Hz
Beat frewuency
Let us consider two sound waves with frequency f1 and f2 such that f1-f2 <10 hz are sounded together then at the starting points can be given as
y1 = asinw1t
y2 = asinw2t
By the principal of superpostion the resultant wave will be,
y = y1+y2
y = asinw1t + asinw2t
y = A sin(pi)(f1+f2) * t
This is a periodic function with amplitude
A = 2acos(pi)(f1-f2) *t
CASE I Loud sound
Loud sound will be heard when value of A becomes maximum i.e when
cos(pi)(f1-f2)*t = 1
cos(pi)(f1-f2)*t = cpsn(pi)
f = n/(f1-f2)
loud sound will be heared at f = 0, 1/(f1-f2),2/(f1-f2)
Beat frequency = (f1-f2) hz
CASEII Soft Sound
Soft sound will be heared when amplitude is minimum i.e when
cos(pi)(f1-f2)*t = cos((2n+1)/2)*pi
f = 1/(f1-f2)
Beat frequency = (f1-f2)Hz
Wave
Wave is the disturbance( means a change in pressure, density or displacement of the particles of the medium about their equilibrium position) produced in the particles of the medium when an energy transfers through the medium.
Transverse wave
If the particle of the medium vibrate along the direction perpendicular to the direction of wave motionm then this tyoe if wave is called a transverse wave.
Longitudinal wave.
If the particle of the medium vibrate along the direction parallel to the direction of wave motion then this type of wave is called longitudinal wave.
Progressive wave
A wave in which the crest and trough( in transvers) or compression and refraction( in lonbgitudinal) travel is called progressive. They vary continiously.
Equation of Progressive wave
Let us consider a progressive wave originating from O travels along the positive x-axis. Let is have frequency f, wavelength λ amplitude a angular velocity V
Since the motion in the medium is simple harmonic motion, the displacement of the particle be given as,
y = a sinwt-----------1
Let us consider a point p at a distance x, from origin O where the wave lags. Let wave lags at P(i.e phase difference) by φ. Then the displacement equation will be
y = asin(wt- φ)-------------2
We know that at distance λ phase difference will be 2 so that the phase difference at distance x will be
φ = 2 x/ λ
equation 2 becomes
y = asin(wt-2 x/ λ)---------------------3
Again,
y = asin(wt-kx)---------------------4
Where k = 2 / λ = wave number
Again using w = 2 / λ we get
y = asin(2 / λ(vt-x))-------------5
These equation 3 4 and 5 are equation of progressive wave.
Particle velocity
The displacement of particle per unit time is called particle velocity we have the displacement of particle when a wave is travelling alon x-axis as
y = asin(wt-kx)---------------------1
Diffrentiating the equation wrt time we get
dy/dx = acos(wt-kx)w
dy/dt = awcos(wt-kx)------------2
Now
Differentiating equation wrt x we get
dy/dx = -akcos(wt-k)-----------3
we have w = 2(pi)f = 2(pi)V/ λ = kV
Equation 2 becomes
dy/dt = akvcos(wt-kx)-------------4
Now from equation 3 and 4 becomes
dy/dt = -v^2dy/dx-------------5
partical velocity = -(wave velocity) * slope of displacement curve at that point
Transverse wave
If the particle of the medium vibrate along the direction perpendicular to the direction of wave motionm then this tyoe if wave is called a transverse wave.
Longitudinal wave.
If the particle of the medium vibrate along the direction parallel to the direction of wave motion then this type of wave is called longitudinal wave.
Progressive wave
A wave in which the crest and trough( in transvers) or compression and refraction( in lonbgitudinal) travel is called progressive. They vary continiously.
Equation of Progressive wave
Let us consider a progressive wave originating from O travels along the positive x-axis. Let is have frequency f, wavelength λ amplitude a angular velocity V
Since the motion in the medium is simple harmonic motion, the displacement of the particle be given as,
y = a sinwt-----------1
Let us consider a point p at a distance x, from origin O where the wave lags. Let wave lags at P(i.e phase difference) by φ. Then the displacement equation will be
y = asin(wt- φ)-------------2
We know that at distance λ phase difference will be 2 so that the phase difference at distance x will be
φ = 2 x/ λ
equation 2 becomes
y = asin(wt-2 x/ λ)---------------------3
Again,
y = asin(wt-kx)---------------------4
Where k = 2 / λ = wave number
Again using w = 2 / λ we get
y = asin(2 / λ(vt-x))-------------5
These equation 3 4 and 5 are equation of progressive wave.
Particle velocity
The displacement of particle per unit time is called particle velocity we have the displacement of particle when a wave is travelling alon x-axis as
y = asin(wt-kx)---------------------1
Diffrentiating the equation wrt time we get
dy/dx = acos(wt-kx)w
dy/dt = awcos(wt-kx)------------2
Now
Differentiating equation wrt x we get
dy/dx = -akcos(wt-k)-----------3
we have w = 2(pi)f = 2(pi)V/ λ = kV
Equation 2 becomes
dy/dt = akvcos(wt-kx)-------------4
Now from equation 3 and 4 becomes
dy/dt = -v^2dy/dx-------------5
partical velocity = -(wave velocity) * slope of displacement curve at that point
Simple Harmonic Motion(SHM)
If in an oscillatory motion the acceleration is directly proportional to displacement and is always directed towards the mean position then it is called SHM
If F be the force applied on the particle and x be the displacement then,
F α –x (-ve sign shows they are opposite in direction)
.-. F = -kx----------1 where k is force constants
If m be the mass of the particle then from Newtons second law of motion.
F = ma-------2
From 1 and 2
ma = -kx
a = -k/m*x
a = -ω^2x----4
This is the equation of simple harmonic motion
Motion of a helical(loded Spring)
Let us consider a negligible helical spring suspended through a rigid support. Now it will attach a mass in its free end then it will be elongated through same distance. let it be L. Then according to Hooke’s law, the restoring force is directly proportional to the elongation produces
F1 α L
F1 = -CL-------------1 (where –ve shows that the force is in the opposite direction of elongation.
Now let the mass be pulled down through a distance x and then realesed. The spring then starts to oscillate. Now elongation on spring will be l+x & restoring force is given as,
F2 = -C(l+x)--------------2
The resulting force is
F = F1 + F2 = C(L+x) + cl = -cx
.-. F = -Cx----------------3
Now
From Newtons second law of motion,
F=ma------------------4
From 3 and 4
ma = -Cx
a = -c/m * x------------5
equation 5 is similar to the equation of SHM. Hence loaded spring executes SHM.
Comparint the equation with SHM we can have
ω^2 = C/m
Now time period of oscillation T = 2 /√(c/m)
Mass attached in two spring
Let us consider a mass m be attached in between two strings s1 and s2 having force constant c1 and c2 respectively.
Let the mass and system be placed in a frictionless surface. Now suppose the mass be pulled toward the right direction through the distance x. Then clearly the spring S1 will be elongated by distance x and spring S2 will be compressed through distance x.
THerefore from Hook’s law restoring force acting on the mass will be
F = -C1x – C2x
F = -(C1+C2)x-------------1
This cause oscillation in mass from Newton’s second law of motion force acting on mass
F=ma-------------2
From 1 and 2
ma = -(C1+C2)x
a = -(C1+C2)/m *x--------3
This equation is similar to equation of SHM. Hence this executes SHM
Simple Pendulum
A point heavy mass suspended with mass less, weightless, extensible string through a rigid support. It is also called ideal pendulum.
Drawbacks of simple pendulum.
It is impossible to have heavy point mass.
It is impossible to have weightless string.
There will be relative motion between the bob and string at extreme position.
Length of pendulum, is difficult to measure exactly accurately.
Compound Pendulum
Compound pendulum is a rigid body of whatever shape capable to oscillate freely about a horizontal axis passing through it.
The point in which pendulum is suspended through rigid suppost is called point of suspension.
The distance between point of suspension and centre of gravity of the pendulum is called the length of pendulum.
Consider a compound pendulum having mass m and length l i.e SG = L. Let pendulum be displace through an angle ϴ. At this instant the weigh of pendulum mf act vertically downward direction. Then the restoring force that tends to bring pendulum torque acting on the body will be,
Ʈ = -mglsin ϴ------------1
Now,
If I be the moment of intertia and
α = a be angular accleration then the restoring torque will be given as
Ʈ = I * α = I * a--------------2
From equation 1 and 2
I * a = -mglsin ϴ
a = -mgl/I * ϴ --------------3
This equation 3 is also similar to equation of SHM. Hence compound also executes SHM.
Point of Suspension and point of oscillation are interchangeable. We have time period for a compound pendulum having length L as
T = 2 √((k^2/L + L)/g)
T = 2 π(√l1 + l)/g-------------1
such that k^2/L = L1
Now, let point of oscillation becomes point of suspension. Then length of pendulum
k^2/L = L1
Hence, time period in this case will be
T1 = 2 π(√k2/l1+l)/g
We have k^2 = LL1
T1 = 2π√LL1+L1/(l/2)
= 2 π√L1 + l)/g-------------2
If F be the force applied on the particle and x be the displacement then,
F α –x (-ve sign shows they are opposite in direction)
.-. F = -kx----------1 where k is force constants
If m be the mass of the particle then from Newtons second law of motion.
F = ma-------2
From 1 and 2
ma = -kx
a = -k/m*x
a = -ω^2x----4
This is the equation of simple harmonic motion
Motion of a helical(loded Spring)
Let us consider a negligible helical spring suspended through a rigid support. Now it will attach a mass in its free end then it will be elongated through same distance. let it be L. Then according to Hooke’s law, the restoring force is directly proportional to the elongation produces
F1 α L
F1 = -CL-------------1 (where –ve shows that the force is in the opposite direction of elongation.
Now let the mass be pulled down through a distance x and then realesed. The spring then starts to oscillate. Now elongation on spring will be l+x & restoring force is given as,
F2 = -C(l+x)--------------2
The resulting force is
F = F1 + F2 = C(L+x) + cl = -cx
.-. F = -Cx----------------3
Now
From Newtons second law of motion,
F=ma------------------4
From 3 and 4
ma = -Cx
a = -c/m * x------------5
equation 5 is similar to the equation of SHM. Hence loaded spring executes SHM.
Comparint the equation with SHM we can have
ω^2 = C/m
Now time period of oscillation T = 2 /√(c/m)
Mass attached in two spring
Let us consider a mass m be attached in between two strings s1 and s2 having force constant c1 and c2 respectively.
Let the mass and system be placed in a frictionless surface. Now suppose the mass be pulled toward the right direction through the distance x. Then clearly the spring S1 will be elongated by distance x and spring S2 will be compressed through distance x.
THerefore from Hook’s law restoring force acting on the mass will be
F = -C1x – C2x
F = -(C1+C2)x-------------1
This cause oscillation in mass from Newton’s second law of motion force acting on mass
F=ma-------------2
From 1 and 2
ma = -(C1+C2)x
a = -(C1+C2)/m *x--------3
This equation is similar to equation of SHM. Hence this executes SHM
Simple Pendulum
A point heavy mass suspended with mass less, weightless, extensible string through a rigid support. It is also called ideal pendulum.
Drawbacks of simple pendulum.
It is impossible to have heavy point mass.
It is impossible to have weightless string.
There will be relative motion between the bob and string at extreme position.
Length of pendulum, is difficult to measure exactly accurately.
Compound Pendulum
Compound pendulum is a rigid body of whatever shape capable to oscillate freely about a horizontal axis passing through it.
The point in which pendulum is suspended through rigid suppost is called point of suspension.
The distance between point of suspension and centre of gravity of the pendulum is called the length of pendulum.
Consider a compound pendulum having mass m and length l i.e SG = L. Let pendulum be displace through an angle ϴ. At this instant the weigh of pendulum mf act vertically downward direction. Then the restoring force that tends to bring pendulum torque acting on the body will be,
Ʈ = -mglsin ϴ------------1
Now,
If I be the moment of intertia and
α = a be angular accleration then the restoring torque will be given as
Ʈ = I * α = I * a--------------2
From equation 1 and 2
I * a = -mglsin ϴ
a = -mgl/I * ϴ --------------3
This equation 3 is also similar to equation of SHM. Hence compound also executes SHM.
Point of Suspension and point of oscillation are interchangeable. We have time period for a compound pendulum having length L as
T = 2 √((k^2/L + L)/g)
T = 2 π(√l1 + l)/g-------------1
such that k^2/L = L1
Now, let point of oscillation becomes point of suspension. Then length of pendulum
k^2/L = L1
Hence, time period in this case will be
T1 = 2 π(√k2/l1+l)/g
We have k^2 = LL1
T1 = 2π√LL1+L1/(l/2)
= 2 π√L1 + l)/g-------------2
Wednesday, June 9, 2010
Structure
Array vs. Structure
1. An array is a collection of related data elements of same type. Structure can have elements of different data types.
2. An array is derived data type whereas a structure is a programmer-defined one.
3. Any array behaves like a built-in data type. All we have to do is to declare an array variable and use it. But in the case of a structure, first we have to design and declare a data structure before the variables of that type are declared anbd used.
Defining a structure
struct tag_name
{
data_type member1;
data_type member2;
};
1. The template is terminated with a semicolon
2. Wgile the entire difinition is considered as a statement, each member is declared independently for its name and type in a separate statement inside the template.
3. The tag name such as book_bank can be used to declare structure variables of the type, later in the program.
Declaring Structure Variables
struct book_bank, book1, book2,book3;
Elements included in declaring structure variable
1. The keyword struct
2. The structure tag name
3. List of variable names separated by commas
4. A terminating semicolon
Accessing Structure members
Members of a structure can be accesed by structure variable( member operator .) variable name
Structur Initialization
Rules for Initialization of Structure
1. We cannot initialize individual members inside the structure template
2. The order of values enclosed in braces must match the order of members in the structure difinition.
3. It is permitted to have a partial initialization. We can initialize only the first few members and leave the remaining blank. The uninitialized member should be only at the end of the list
4. The uninitialized members will be assigned default value as zero for intiger and #0 for character string
Unions
Major distinction between structure and union is in terms of storage. In structure, each member has its own storage location, whereas all the members of a union use the same location. This implies that, although a union may contain many member of different types, it can handle only one member at a time
1. An array is a collection of related data elements of same type. Structure can have elements of different data types.
2. An array is derived data type whereas a structure is a programmer-defined one.
3. Any array behaves like a built-in data type. All we have to do is to declare an array variable and use it. But in the case of a structure, first we have to design and declare a data structure before the variables of that type are declared anbd used.
Defining a structure
struct tag_name
{
data_type member1;
data_type member2;
};
1. The template is terminated with a semicolon
2. Wgile the entire difinition is considered as a statement, each member is declared independently for its name and type in a separate statement inside the template.
3. The tag name such as book_bank can be used to declare structure variables of the type, later in the program.
Declaring Structure Variables
struct book_bank, book1, book2,book3;
Elements included in declaring structure variable
1. The keyword struct
2. The structure tag name
3. List of variable names separated by commas
4. A terminating semicolon
Accessing Structure members
Members of a structure can be accesed by structure variable( member operator .) variable name
Structur Initialization
Rules for Initialization of Structure
1. We cannot initialize individual members inside the structure template
2. The order of values enclosed in braces must match the order of members in the structure difinition.
3. It is permitted to have a partial initialization. We can initialize only the first few members and leave the remaining blank. The uninitialized member should be only at the end of the list
4. The uninitialized members will be assigned default value as zero for intiger and #0 for character string
Unions
Major distinction between structure and union is in terms of storage. In structure, each member has its own storage location, whereas all the members of a union use the same location. This implies that, although a union may contain many member of different types, it can handle only one member at a time
Pointers
Why Pointers
1. Pointers are more efficient in handling arrays and data tables.
2. Pointers can be used to return multiple values from a function via function arguments.
3.Pointers permit references to functions and thereby facilitating passing of function s argument to another function.
4.The use of pointer arrays to character strings results in saving of data storage space in memory.
5.Pointers allow C to support dynamic memory managment.
6. Pointers reduce length and complexity of programs
7. They reduce the execution speed and thus reduce the program execution time.
Pointer Constant Pointer values Pointer Variable
Memory addresses within a computer are referred to as pointer constants. We can not change them; we can only use them to store data values. They are like house numbers.
We cannot save the value of a memory address directly. We can only obtain the value through the variable stored there using the address operator &. The value thus obtained is known as pointer value. The pointer value(i.e. the address of a variable) may change from one run of the program to another.
Once we have a pointer value, it can be stored into another variable. The variable that contains a pointer value is called pointer variable.
Rules of Pointer Operations
1. A pointer variable can be assigned the address of another variable.
2. A pointer variable can be assigned the values of another pointer variable.
3. A pointer variable can be initialized with NULL or zero value.
4. A pointer variable can be pre-fixed or post-fixed with increment or decrement operators
5. An integer value may be added or subtracted from a pointer variable.
6. When two pointers point to the same array, one pointer variable can b e subtracted from another.
7. When two pointers points to the objects of the same data types, they can be compared using relational operators.
8. A pointer variable can not be muktiplied by a constant
9. Two pointer variables can not be added
10. A value can not be assigned to an arbitary address
1. Pointers are more efficient in handling arrays and data tables.
2. Pointers can be used to return multiple values from a function via function arguments.
3.Pointers permit references to functions and thereby facilitating passing of function s argument to another function.
4.The use of pointer arrays to character strings results in saving of data storage space in memory.
5.Pointers allow C to support dynamic memory managment.
6. Pointers reduce length and complexity of programs
7. They reduce the execution speed and thus reduce the program execution time.
Pointer Constant Pointer values Pointer Variable
Memory addresses within a computer are referred to as pointer constants. We can not change them; we can only use them to store data values. They are like house numbers.
We cannot save the value of a memory address directly. We can only obtain the value through the variable stored there using the address operator &. The value thus obtained is known as pointer value. The pointer value(i.e. the address of a variable) may change from one run of the program to another.
Once we have a pointer value, it can be stored into another variable. The variable that contains a pointer value is called pointer variable.
Rules of Pointer Operations
1. A pointer variable can be assigned the address of another variable.
2. A pointer variable can be assigned the values of another pointer variable.
3. A pointer variable can be initialized with NULL or zero value.
4. A pointer variable can be pre-fixed or post-fixed with increment or decrement operators
5. An integer value may be added or subtracted from a pointer variable.
6. When two pointers point to the same array, one pointer variable can b e subtracted from another.
7. When two pointers points to the objects of the same data types, they can be compared using relational operators.
8. A pointer variable can not be muktiplied by a constant
9. Two pointer variables can not be added
10. A value can not be assigned to an arbitary address
Decision Making and Looping
THE WHILE STATEMENT
while (test condition)
{
body of the loop
}
The while is an entry-controlled loop statement. The test-condition is evaluated and if the condition is true, then the body of the loop is executed. After execution of the body, the test-condition is once again evaluated and if it is true, the body is executed once again. This process of repeated execution of the body continues until the test-condition finally becomes false and the control is transferred out of the loop. On exit the program continues with the statement immediately after the body of the loop.
THE DO STATEMENT
do
{
body of the loop
}
while (test-condition);
The di statement is an exit-controlled loop. On reaching the do statement, the program proceeds to evaluate the body loop first. At the end of the loop, the test-condition in the while statement is evaluated. If the condition is true, the program continues to evaluate the body of the loop once again. This process continues as long as the condition is true. When the condition become false, the loop will be terminated and the control goes to the statement that appears immediately after the while statement. As it is a exit controlled loop and therefore the body of the loop is always executed at least once.
THE FOR STATEMENT
The for loop is another entry-controlled loop that provides a more consise loop control structure. The general form of the for loop is
for (initialization)
{
body of the loop
}
The execution of the for statement is as follows:
1.Initialization of the control variables is done first, using assignment statements such as i=1 and count = 0. The varianles i and count are known as loop-controlled variables.
2.The value of the control variable is tested using the test-condition. The test-condition is a relational expression, such as i<10 that determines when the loop will exit. If the condition is true, the body of the loop is executed; otherwise the loop is terminated and the execution continues with the statement that immediately follows the loop.
3. When the body of the loop is executed, the control is transferred back to the for statement after evaluating the last statement in the loop. Now the control variable is incremented using an assignment statement such as i=i+1 and the new value of the control variable is again tested to see whether is satisfies the loop condition. if the condition is satisfied, the body of the loop is again executed. This process continues till the value of the control variable fails to satisfy the test-condition.
Break Statement
When a break statement is encountered inside a loop, the loop is immediately exited and the program continues with the statement immediately following the loop. When the loop are nested, the break only exit from the loop containing it. That is, the break exit only a single loop.
Continue Statement
The continue statement tells the compiler, SKIP THE FOLLOWING STATEMENTS AND CONTINUE WITH THE NEXT ITERATION its systax is simply continue;
while (test condition)
{
body of the loop
}
The while is an entry-controlled loop statement. The test-condition is evaluated and if the condition is true, then the body of the loop is executed. After execution of the body, the test-condition is once again evaluated and if it is true, the body is executed once again. This process of repeated execution of the body continues until the test-condition finally becomes false and the control is transferred out of the loop. On exit the program continues with the statement immediately after the body of the loop.
THE DO STATEMENT
do
{
body of the loop
}
while (test-condition);
The di statement is an exit-controlled loop. On reaching the do statement, the program proceeds to evaluate the body loop first. At the end of the loop, the test-condition in the while statement is evaluated. If the condition is true, the program continues to evaluate the body of the loop once again. This process continues as long as the condition is true. When the condition become false, the loop will be terminated and the control goes to the statement that appears immediately after the while statement. As it is a exit controlled loop and therefore the body of the loop is always executed at least once.
THE FOR STATEMENT
The for loop is another entry-controlled loop that provides a more consise loop control structure. The general form of the for loop is
for (initialization)
{
body of the loop
}
The execution of the for statement is as follows:
1.Initialization of the control variables is done first, using assignment statements such as i=1 and count = 0. The varianles i and count are known as loop-controlled variables.
2.The value of the control variable is tested using the test-condition. The test-condition is a relational expression, such as i<10 that determines when the loop will exit. If the condition is true, the body of the loop is executed; otherwise the loop is terminated and the execution continues with the statement that immediately follows the loop.
3. When the body of the loop is executed, the control is transferred back to the for statement after evaluating the last statement in the loop. Now the control variable is incremented using an assignment statement such as i=i+1 and the new value of the control variable is again tested to see whether is satisfies the loop condition. if the condition is satisfied, the body of the loop is again executed. This process continues till the value of the control variable fails to satisfy the test-condition.
Break Statement
When a break statement is encountered inside a loop, the loop is immediately exited and the program continues with the statement immediately following the loop. When the loop are nested, the break only exit from the loop containing it. That is, the break exit only a single loop.
Continue Statement
The continue statement tells the compiler, SKIP THE FOLLOWING STATEMENTS AND CONTINUE WITH THE NEXT ITERATION its systax is simply continue;
Functions
Need for user defined function
1. It facilitates top-down modular programming.
2. The length of a source program can be reduced by using functions at appropriate places.
3. It is easy to lacate and isolate a faulty function for further investigations.
4. A function may be used by many other program
Definition of function
A function definition, also known as function implementation shall include the following elements;
1.function name;
2.function type;
3.list of parameteres;
4.local variable decleration;
5.function statement;
6.a return statement;
General format
function_type function_name(parameter list)
{
local variable declaration;
executable statement1;
executable statement2;
.....
......
return statement;
}
Category of function
No arguments and no return values
Argumentd but no return values
Argumentd with return values
No arguments but returns a value
Function that returns multiple value
The scope visibility and lifetime of variables
Types of variables
1.Automatic
They created when the function is called and destroyed automatically when the function is exited.
They are private (local) to the function in which they are declared. So also called local variable.
It is assign to a variable by default if no storage class is specified.
2.External
They are both alive and active through out the program.
Also known as global variables.
They are declared outside of a function
3. Static
The value of the static variables presists until the end of the program.
Key word static can be used to declare the variable
Internal static variable are similar to automatic variable
Static variables can be used to retain values between function calls
4.Register
Variables kept in the machine's registers.
Variables that needs much faster access than that given by memory access.
Most compiler allow only int or char variables to be placed in the register.
Scope Visibility and Lifetime
Scope
The region of a program in which a variable is available for use.
Visibility
The Program's ability to access variable from the memory.
Lifetime
The lifetime of a variable is the duration of time in which a variable exists in the memory during execution.
Rules of use
1. The scope of a global variable is the entire program file.
2. The scope of a local variable begins at point of declaration and ends at the end of the block or function in which it is declared.
3. The scope of a formal function argument is its own function.
4. The lifetime of an auto variable declared in main is the entire program execution time, although its scope is oonly in the main function.
5. The life of an auto variable declared in a function ends when the function is exited.
6. A static local variable, although its scope is limited to its function, its lifetime extends till the end of program execution.
7. All variables have visibility in their scope, provided that are not declared again.
8. If a variable is redeclared within its scope again, it loses its visibility in the scope of the redeclared variable.
1. It facilitates top-down modular programming.
2. The length of a source program can be reduced by using functions at appropriate places.
3. It is easy to lacate and isolate a faulty function for further investigations.
4. A function may be used by many other program
Definition of function
A function definition, also known as function implementation shall include the following elements;
1.function name;
2.function type;
3.list of parameteres;
4.local variable decleration;
5.function statement;
6.a return statement;
General format
function_type function_name(parameter list)
{
local variable declaration;
executable statement1;
executable statement2;
.....
......
return statement;
}
Category of function
No arguments and no return values
Argumentd but no return values
Argumentd with return values
No arguments but returns a value
Function that returns multiple value
The scope visibility and lifetime of variables
Types of variables
1.Automatic
They created when the function is called and destroyed automatically when the function is exited.
They are private (local) to the function in which they are declared. So also called local variable.
It is assign to a variable by default if no storage class is specified.
2.External
They are both alive and active through out the program.
Also known as global variables.
They are declared outside of a function
3. Static
The value of the static variables presists until the end of the program.
Key word static can be used to declare the variable
Internal static variable are similar to automatic variable
Static variables can be used to retain values between function calls
4.Register
Variables kept in the machine's registers.
Variables that needs much faster access than that given by memory access.
Most compiler allow only int or char variables to be placed in the register.
Scope Visibility and Lifetime
Scope
The region of a program in which a variable is available for use.
Visibility
The Program's ability to access variable from the memory.
Lifetime
The lifetime of a variable is the duration of time in which a variable exists in the memory during execution.
Rules of use
1. The scope of a global variable is the entire program file.
2. The scope of a local variable begins at point of declaration and ends at the end of the block or function in which it is declared.
3. The scope of a formal function argument is its own function.
4. The lifetime of an auto variable declared in main is the entire program execution time, although its scope is oonly in the main function.
5. The life of an auto variable declared in a function ends when the function is exited.
6. A static local variable, although its scope is limited to its function, its lifetime extends till the end of program execution.
7. All variables have visibility in their scope, provided that are not declared again.
8. If a variable is redeclared within its scope again, it loses its visibility in the scope of the redeclared variable.
Decision Making and Branching
If statement
The if ststement is a powerful decision-making ststement and is used to control the flow of execution of statements. It is basically a two-way decision statement and is used in conjuction with an expression
SIMPLE IF STATEMENT
if (test expression)
{
statement-block;
}
statement-x;
THE IF......ELSE STATEMENT
if (test expression)
{
True-block statement(s)
}
else
{
False-block statement(s)
}
statement-x
THE ELSE IF LADER
if (condition 1)
statement-1;
else if (condition 2)
statement-2;
else if (condition2)
statement-3;
else
default-ststement;
statement-x;
THE SWITCH STATEMENT
The switch statements tests the value of a given variable (or expression) against a list of case values and when a match is found , a block of statements associated with that case is executed. The general form of the switch statement is
switch (expression)
{
case value-1:
block-1
break;
case value-2:
block-2
break;
default:
default-block
break;
}
statemeny-x;
Rules for switch Statement
1. The switch expression must be an integral type.
2. Case labels must be constants or constant expressions.
3. Case label must be unique.
4. It is permitted to nest switch statements
5. There can be at most one default label
6.The default may be placed anywhere but uually placed at the end
The if ststement is a powerful decision-making ststement and is used to control the flow of execution of statements. It is basically a two-way decision statement and is used in conjuction with an expression
SIMPLE IF STATEMENT
if (test expression)
{
statement-block;
}
statement-x;
THE IF......ELSE STATEMENT
if (test expression)
{
True-block statement(s)
}
else
{
False-block statement(s)
}
statement-x
THE ELSE IF LADER
if (condition 1)
statement-1;
else if (condition 2)
statement-2;
else if (condition2)
statement-3;
else
default-ststement;
statement-x;
THE SWITCH STATEMENT
The switch statements tests the value of a given variable (or expression) against a list of case values and when a match is found , a block of statements associated with that case is executed. The general form of the switch statement is
switch (expression)
{
case value-1:
block-1
break;
case value-2:
block-2
break;
default:
default-block
break;
}
statemeny-x;
Rules for switch Statement
1. The switch expression must be an integral type.
2. Case labels must be constants or constant expressions.
3. Case label must be unique.
4. It is permitted to nest switch statements
5. There can be at most one default label
6.The default may be placed anywhere but uually placed at the end
Constants, Variables
Keywords
Keywords serve as the building blocks for program statements. All keywords have fixed meanings and these meanings cannot be changed. All keywords must be written in lower case. Example break else long switch void
Identifiers
Identifiers refer to the names of variables, functions and arrays. Thses are user-defined names and consist of a sequence of letters and digits, with a letter as a first character. Both uppercase and lowercase letters are permitted, although lowercase letters are commonly used.
Rules for Identifiers
1. First Character must be an alphabet(or underscore)
2.Must consist of only letters, digits or underscore.
3.Only first 31 character are significant.
4.Cannot use a keyword
5.Must not contain white space.
Constants
Constants in C refer to fixed values that do not change during the execution of a Program.
Types of Constants
Integer Constants refers to sequence of digits
Real Constants referes to number containing fractional parts like 17.548
String Character Constants contains a single character en-closed within a pait of single quote marks.
String Constants is a sequence of characters enclosed in double quotes.
Backslash Character Constants Escape Sequence
\a----audible alert
\b backspace
\f form feed
\n new line
\r carriage return
\\ backshash
\0 null
Variable
A variable is a data name that may be used to store a data value. Unlike constants that remain unchanged during the execution of a program, a variable may take different values at different time during execution.
Points to be noted while having variable
1. They mush begin with a letter. Some system permits underscore as a first character
2. A length of only 31 character are recognize by ANSI standared
3.Uppercase and Lowercase are significant
4.It should not be a keyword
5.White Space is not allowed
Data Types
Integer Types
Integers are whole numbers with a machine dependent range of values. C has three classes of integer storage namely short int, int and long int. All of this data types have signed and unsigned form
Floating Point types
Floating point number represents a real number with 6 digits precision. Floating point numbers are denoted by the key word float. When the accuracy of float is insufficient we can use the double to define the number. The double is same as float but with longer precision.
Void Type
Using void data type we can specify the data we can specify the type of function. It is a good practice to avoid functions that doesnot return any value to the calling function.
Character Type
A single character can be defined as a character type of data. Characters are usually stored in 8 bits of internal storage
Keywords serve as the building blocks for program statements. All keywords have fixed meanings and these meanings cannot be changed. All keywords must be written in lower case. Example break else long switch void
Identifiers
Identifiers refer to the names of variables, functions and arrays. Thses are user-defined names and consist of a sequence of letters and digits, with a letter as a first character. Both uppercase and lowercase letters are permitted, although lowercase letters are commonly used.
Rules for Identifiers
1. First Character must be an alphabet(or underscore)
2.Must consist of only letters, digits or underscore.
3.Only first 31 character are significant.
4.Cannot use a keyword
5.Must not contain white space.
Constants
Constants in C refer to fixed values that do not change during the execution of a Program.
Types of Constants
Integer Constants refers to sequence of digits
Real Constants referes to number containing fractional parts like 17.548
String Character Constants contains a single character en-closed within a pait of single quote marks.
String Constants is a sequence of characters enclosed in double quotes.
Backslash Character Constants Escape Sequence
\a----audible alert
\b backspace
\f form feed
\n new line
\r carriage return
\\ backshash
\0 null
Variable
A variable is a data name that may be used to store a data value. Unlike constants that remain unchanged during the execution of a program, a variable may take different values at different time during execution.
Points to be noted while having variable
1. They mush begin with a letter. Some system permits underscore as a first character
2. A length of only 31 character are recognize by ANSI standared
3.Uppercase and Lowercase are significant
4.It should not be a keyword
5.White Space is not allowed
Data Types
Integer Types
Integers are whole numbers with a machine dependent range of values. C has three classes of integer storage namely short int, int and long int. All of this data types have signed and unsigned form
Floating Point types
Floating point number represents a real number with 6 digits precision. Floating point numbers are denoted by the key word float. When the accuracy of float is insufficient we can use the double to define the number. The double is same as float but with longer precision.
Void Type
Using void data type we can specify the data we can specify the type of function. It is a good practice to avoid functions that doesnot return any value to the calling function.
Character Type
A single character can be defined as a character type of data. Characters are usually stored in 8 bits of internal storage
Friday, June 4, 2010
Magic Square
Magic Square
3*3
816
357
492
4*4
16020313
05111008
09070612
04141501
5*5
2306190215
1018011422
1705132109
0412250816
1124072003
3*3
816
357
492
4*4
16020313
05111008
09070612
04141501
5*5
2306190215
1018011422
1705132109
0412250816
1124072003
Crypto
Cryptos
SEVEN
+EIGHT
TWELVE
85254
+50671
135925
DONALD
+GERALD
ROBERT
526485
197485
723970
SEND
+MORE
MONEY
9567
+1085
10652
LETS
+WAVE
LATER
1567
+9085
10652
WRONG
WRONG
RIGHT
12734
12734
25468
(ATOM)^(1/2) = A+TO+M
(1296)^(1/2) = 1+29+6
FIVE
ONE
FOUR
1032
652
1684
SEVEN
+EIGHT
TWELVE
85254
+50671
135925
DONALD
+GERALD
ROBERT
526485
197485
723970
SEND
+MORE
MONEY
9567
+1085
10652
LETS
+WAVE
LATER
1567
+9085
10652
WRONG
WRONG
RIGHT
12734
12734
25468
(ATOM)^(1/2) = A+TO+M
(1296)^(1/2) = 1+29+6
FIVE
ONE
FOUR
1032
652
1684
Pattern
Determine the pattern to prove the identity
1 = 1
3+5 = 8
7+9+11 = 27
According to the given pattern in question
the general pattern is
(n^2-n+1) + (n^2-n+3) + (n^2-n+5).....(n^2-n+(2n-1)) = n^3
Here we need tp proof LHS = RHS
left hand side has n summands
n*n^2 - n*n + (1+3+5+.....2)
i.e n mumbers of n^2+1 and sum of odd numners only
Now, 1+3+5+.....(2n-1) equals to
(2*1-1)+(2*2-1)+(2*3-1)+....(2n+1)
2(1+2+3+4+....n)-1(1+1+1+1+1...)
2(n^2-n)/2 - n = n^2
n*n^2-n*n + n^2
n^3 = RHS
Pattern
1 = 1
2+3+4 = 1+8
5+6+7+8+9 = 8 + 27
1^3 = 1^3
(1^2+1) + (1^2+2) + 2^2 = 1^3 + 2^3
(2^2+1) + (2^2+2) + (2^2+3) + (2^2+4) + 3^2 = 2^3 + 3^3
((n-1)^2+1) + ((n-1)^2+2) + ((n-1)^2+3) + ((n-1)^2+4)+n^2 = (n-1)^3 + n^3
sum of integer upto n^2 - sum of integer upto (n-1)^2
(1+2+3+.....9)-(1.2...4) = (5+6+7+8+9)
[n^2*(n^2+1)]/2 =[(n-1)^2(n+1)^2+1]/2
Pattern
1=1
1-4= -(1+2)
1-4+9 = +(1+2+3)
1-4+9-16 = -(1+2+3+4)
general pattern is
1^2 - 2^2 + 3^2 - 4^2 + h^2 .......[(-1)^(n+1) - n^2] = (-1)^(n+1)(1+2+3...n)
For even
(1^2-2^2) + (3^2-4^2)+......[(n-1)^2 - n^2]
(1-2)(1+2) + (3-4)(3+4)......-1(n-1)+n
-1(1+2+3+4+.....(n-1)+n)
For odd
(1^2-2^2) + (3^2-4^2)+......[(n-2)^2-(n-1^2)] + n^2
(1^2-2^2) + (3^2-4^2)+......(n-2-n+1)(n-2+n-1) + n^2
-1(1+2)-1(3+4)+ -1[(n-2)+(n-1)^2] + n^2
-1{(1+2+3+4...(n-2)+(n-1)} + n^2
-n(n+1)/2 + n(n+1)
n(n+1)/2
1+2+3+4...n
1 = 1
3+5 = 8
7+9+11 = 27
According to the given pattern in question
the general pattern is
(n^2-n+1) + (n^2-n+3) + (n^2-n+5).....(n^2-n+(2n-1)) = n^3
Here we need tp proof LHS = RHS
left hand side has n summands
n*n^2 - n*n + (1+3+5+.....2)
i.e n mumbers of n^2+1 and sum of odd numners only
Now, 1+3+5+.....(2n-1) equals to
(2*1-1)+(2*2-1)+(2*3-1)+....(2n+1)
2(1+2+3+4+....n)-1(1+1+1+1+1...)
2(n^2-n)/2 - n = n^2
n*n^2-n*n + n^2
n^3 = RHS
Pattern
1 = 1
2+3+4 = 1+8
5+6+7+8+9 = 8 + 27
1^3 = 1^3
(1^2+1) + (1^2+2) + 2^2 = 1^3 + 2^3
(2^2+1) + (2^2+2) + (2^2+3) + (2^2+4) + 3^2 = 2^3 + 3^3
((n-1)^2+1) + ((n-1)^2+2) + ((n-1)^2+3) + ((n-1)^2+4)+n^2 = (n-1)^3 + n^3
sum of integer upto n^2 - sum of integer upto (n-1)^2
(1+2+3+.....9)-(1.2...4) = (5+6+7+8+9)
[n^2*(n^2+1)]/2 =[(n-1)^2(n+1)^2+1]/2
Pattern
1=1
1-4= -(1+2)
1-4+9 = +(1+2+3)
1-4+9-16 = -(1+2+3+4)
general pattern is
1^2 - 2^2 + 3^2 - 4^2 + h^2 .......[(-1)^(n+1) - n^2] = (-1)^(n+1)(1+2+3...n)
For even
(1^2-2^2) + (3^2-4^2)+......[(n-1)^2 - n^2]
(1-2)(1+2) + (3-4)(3+4)......-1(n-1)+n
-1(1+2+3+4+.....(n-1)+n)
For odd
(1^2-2^2) + (3^2-4^2)+......[(n-2)^2-(n-1^2)] + n^2
(1^2-2^2) + (3^2-4^2)+......(n-2-n+1)(n-2+n-1) + n^2
-1(1+2)-1(3+4)+ -1[(n-2)+(n-1)^2] + n^2
-1{(1+2+3+4...(n-2)+(n-1)} + n^2
-n(n+1)/2 + n(n+1)
n(n+1)/2
1+2+3+4...n
k positive integers
Find the sum of k positive integers
1+2+3+4+........+k = s(k)
1+2+3+4+(k-1)+k+(k+1) = s(k+1)
using linear equation
(k+1)^2 = K^2 + 2K + 1
(k+1)^2 - k^2 = 2K + 1-----------1
When k = 1
2^2-1^2 = 2*1 +1
When k = 2
3^2-2^2 = 2*2 +1
When k = 3
4^2-3^2 = 2*3 +1
When k = 4
5^2-4^2 = 2*4 +1
When k = k
(k+1)^2 - k^2 = 2K + 1
Now adding both sides
(2^2-1^2)+(3^2-2^2)+.........(k^2-(k-1)^2) + (k+1)^2 - k^2 = 2(1+2+3+4....k) + (1+1+1+1+ to k)
(k+1)^2 - 1^2 = 2s(k) + k
K^2 + 2k +1 -1 = 2s(k) + k
s(k) = (K^2 + k)/2 - k(k+1)/2
Find the sum of k^2 positive integers
1^2 + 2^2 + 3^2 + 4^2 +.......k^2 = s(k^2)
1^2 + 2^2 + 3^2 + 4^2 + (k-1)^2 k^2 + (k+1)^2 = s((k+1)^2)
using equation.
k^3 - (k-1)^3 = 3k^2 - 3K + 1
put k = 1,1,2,3 ....... respectively
0^3 - (0-1)^3 = 3*0^2 - 3*0 + 1
1^3 - (1-1)^3 = 3*1^2 - 3*1 + 1
2^3 - (2-1)^3 = 3*2^2 - 3*2 + 1
..
..
..
k^3 - (k-1)^3 = 3k^2 - 3*k + 1
-------------------------------Adding both sides
k^3 = 3(0^21^2 + 2^2 + 3^2 + 4^2 + k^2) - 3(1+2+3.....k) + k
k^3 = 3s(k) - 3k(k+1)/2 + k
k^3 + 3k(k+1)/2 - k = 3s(k)
s(k) = K^3/3 + (3k^2 + 3k)/6 - k/3
(2k^3 + 3k^2 + 3K - 2k)/6
.....
.....
s(k) = (k(2K+1)(k+1))/6
Claculate the sum of first k odd positive integer
1+3+5+7+......(2k-1)
First adding all numbers from 1 + (2k-1) and subtracting even numbers
1+2+3+4+.....(2k-1)-[2+4+6+....2k-2]
((2k-1)^2 +(2k-1))/2 - 2[(k-1)^2)+(k-1)]/2
[(2k-1).2k]/2 - 2/2(k-1) * k
2k^2 - k^2
k^2
1+2+3+4+........+k = s(k)
1+2+3+4+(k-1)+k+(k+1) = s(k+1)
using linear equation
(k+1)^2 = K^2 + 2K + 1
(k+1)^2 - k^2 = 2K + 1-----------1
When k = 1
2^2-1^2 = 2*1 +1
When k = 2
3^2-2^2 = 2*2 +1
When k = 3
4^2-3^2 = 2*3 +1
When k = 4
5^2-4^2 = 2*4 +1
When k = k
(k+1)^2 - k^2 = 2K + 1
Now adding both sides
(2^2-1^2)+(3^2-2^2)+.........(k^2-(k-1)^2) + (k+1)^2 - k^2 = 2(1+2+3+4....k) + (1+1+1+1+ to k)
(k+1)^2 - 1^2 = 2s(k) + k
K^2 + 2k +1 -1 = 2s(k) + k
s(k) = (K^2 + k)/2 - k(k+1)/2
Find the sum of k^2 positive integers
1^2 + 2^2 + 3^2 + 4^2 +.......k^2 = s(k^2)
1^2 + 2^2 + 3^2 + 4^2 + (k-1)^2 k^2 + (k+1)^2 = s((k+1)^2)
using equation.
k^3 - (k-1)^3 = 3k^2 - 3K + 1
put k = 1,1,2,3 ....... respectively
0^3 - (0-1)^3 = 3*0^2 - 3*0 + 1
1^3 - (1-1)^3 = 3*1^2 - 3*1 + 1
2^3 - (2-1)^3 = 3*2^2 - 3*2 + 1
..
..
..
k^3 - (k-1)^3 = 3k^2 - 3*k + 1
-------------------------------Adding both sides
k^3 = 3(0^21^2 + 2^2 + 3^2 + 4^2 + k^2) - 3(1+2+3.....k) + k
k^3 = 3s(k) - 3k(k+1)/2 + k
k^3 + 3k(k+1)/2 - k = 3s(k)
s(k) = K^3/3 + (3k^2 + 3k)/6 - k/3
(2k^3 + 3k^2 + 3K - 2k)/6
.....
.....
s(k) = (k(2K+1)(k+1))/6
Claculate the sum of first k odd positive integer
1+3+5+7+......(2k-1)
First adding all numbers from 1 + (2k-1) and subtracting even numbers
1+2+3+4+.....(2k-1)-[2+4+6+....2k-2]
((2k-1)^2 +(2k-1))/2 - 2[(k-1)^2)+(k-1)]/2
[(2k-1).2k]/2 - 2/2(k-1) * k
2k^2 - k^2
k^2
Tuesday, June 1, 2010
Road Foundation
1.Preperation
- Survey to determine
* line of road
* Amount of excavation and refilling likely to be needed
* Amount of surfacing material
* Quality of soil
2. Factors to take into account
Soil should be permeable for good drainage and easily compacted to good bearing capacity otherwise the road surface will not be durable
3. Process
Site should be excavated unsuitable soil should be replaced various machine are available. It depends on the depth of the cut and how much space there is at the site.
Final compacting is done with a roller. This complets the sub-grade
Then sub-base layer is added should also consist of material that drains easily
This layer must be compacted carefully so that its density is uniform all over it
The layer must be sealed
- Survey to determine
* line of road
* Amount of excavation and refilling likely to be needed
* Amount of surfacing material
* Quality of soil
2. Factors to take into account
Soil should be permeable for good drainage and easily compacted to good bearing capacity otherwise the road surface will not be durable
3. Process
Site should be excavated unsuitable soil should be replaced various machine are available. It depends on the depth of the cut and how much space there is at the site.
Final compacting is done with a roller. This complets the sub-grade
Then sub-base layer is added should also consist of material that drains easily
This layer must be compacted carefully so that its density is uniform all over it
The layer must be sealed
The use and Mis-use of Science
1)) An invention can be good or bad
the motorcar--facilitates business and gives harmless enjoyment/////strew the roads with dead
the radio--link the world together in a moment////instrument of lying
the cinema--means of instruction and recreation/////channel of vulgarity and false values
aeroplanes---makes travel rapid and easy/////weapon of
the motorcar--facilitates business and gives harmless enjoyment/////strew the roads with dead
the radio--link the world together in a moment////instrument of lying
the cinema--means of instruction and recreation/////channel of vulgarity and false values
aeroplanes---makes travel rapid and easy/////weapon of
Knowledge and Wisdom
Knowledge and wisdom are two entirely different thing though they sound alike. We can find a lot of difference between knowledge and wisdom. Knowledge is defined as the perception or learning of various things we go through but wisdom is actually the accumulated knowledge where we use such knowledge and experience with common sense. Knowledge in some field helps us to think deeply and come up with some wonderfull discovery in that particular fiels only but wisdom helps us to think rationally on what impact can that discovery has to the other field associated with it along with these discoveries. Wisdom allows us to think for the mankind and also motivates ourselves to make continual approaches towards impartiallywhich knowledge simply can not do. Sometimes resulting in the destruction. But wisdom never takes the mind towards it. These are the difference we can find between knowledge and wisdom.
Customs
Two definition of culture
The first definition has given is from an anthropologist prespectives. He says that culture is the total life way of a people. The social legacy the indivudual acquires from his group. In other words, cl#ulture can be regared as that part of the environment that is the creation of man. The first definition has a wider meaning because the concept od culture needs a clear explanation of an individuals past experience. past experience of other people in the society has. Every activity from birth to death and total relation with the patterns which is not his own making. It is bacuse culture is a man made phenomenon and much of derived from the culture.
In second definition of culture is that a person who know about literature, philosophy history and fine arts is a cultured person. Culture denotes one's knowledge of language pther than his own. It is not a man made environment and it it limits the culture into a kind of mental ability. It also suggent that culture is the knoeledge about picaso.
At bottom all human being are very much alike
By asserting that at bottom all human being are very much alike he is not contradicting himself. rather he is useing a technical device in his persuasive essay. As an anthropologist kluckhohn does not conclude his essay with definition of cultur and some examples of cultural differences but he begins his essay with rhetorical question defines culture and develops his points with some supportive arguments to answer the questions and examiners human being's behaviours and character culturally and at last biologically too. Thus his essay eventually turns out to be through examination of human beings- their biological properties way of living and how they are shared by a certain pattern created by themselves.
Until paragraph 8 the author argues and draws examples and anologies to show that human being are different because of their cultural difference. It is bacasue they are btought up in different way, every cultural f#group has differemt customs such as chinese, japanese, indians Europeans, americans Africans and so on. These are some obvious directives or diversities among human being because of diverse cultural heritage.
However in the concluding pe#aragraph hr treats human being as one species if we observe human being's biological peoperties. the sum biologycal equipments and the inevitable of biology are found. All man undergo the same biologicaal life experience as birth , helplessness illness oldage death irrespective of cultural gropus their biological needed and same social behaviours such as marriage ceremony life crisis rite and incest taboos are alike.
The first definition has given is from an anthropologist prespectives. He says that culture is the total life way of a people. The social legacy the indivudual acquires from his group. In other words, cl#ulture can be regared as that part of the environment that is the creation of man. The first definition has a wider meaning because the concept od culture needs a clear explanation of an individuals past experience. past experience of other people in the society has. Every activity from birth to death and total relation with the patterns which is not his own making. It is bacuse culture is a man made phenomenon and much of derived from the culture.
In second definition of culture is that a person who know about literature, philosophy history and fine arts is a cultured person. Culture denotes one's knowledge of language pther than his own. It is not a man made environment and it it limits the culture into a kind of mental ability. It also suggent that culture is the knoeledge about picaso.
At bottom all human being are very much alike
By asserting that at bottom all human being are very much alike he is not contradicting himself. rather he is useing a technical device in his persuasive essay. As an anthropologist kluckhohn does not conclude his essay with definition of cultur and some examples of cultural differences but he begins his essay with rhetorical question defines culture and develops his points with some supportive arguments to answer the questions and examiners human being's behaviours and character culturally and at last biologically too. Thus his essay eventually turns out to be through examination of human beings- their biological properties way of living and how they are shared by a certain pattern created by themselves.
Until paragraph 8 the author argues and draws examples and anologies to show that human being are different because of their cultural difference. It is bacasue they are btought up in different way, every cultural f#group has differemt customs such as chinese, japanese, indians Europeans, americans Africans and so on. These are some obvious directives or diversities among human being because of diverse cultural heritage.
However in the concluding pe#aragraph hr treats human being as one species if we observe human being's biological peoperties. the sum biologycal equipments and the inevitable of biology are found. All man undergo the same biologicaal life experience as birth , helplessness illness oldage death irrespective of cultural gropus their biological needed and same social behaviours such as marriage ceremony life crisis rite and incest taboos are alike.
Beauty
Conventional attitude about beauty that sontag seeks to discredit?
The conventionally used beauty signifiesthe attraction of physical appearance. Beauty has been distinguished as inner beauty(character intellect and vision) and outer beauty(facial attraction, physique or sense of proportion of the body) which is attributed to women only. The author seeks to discredit this outer beauty because it has devalued the concept of beauty which signified the wholeness, the excellence and the virtue of human during the classical time.
Sontag wants to revive the old Greek concept of inner beauty. Soctates diciples had observed the inner beauty in their teacher although he was ugly looking person.
If beauty is a source of power, why does sontag object to wone’s striving to attain it?
Beauty is a form of power. She always looked at with suspicious eye even if she has good rise in work, politics, law, medicine, business or whatever. She is always pressure to confess that she stills works at being attractive. This power is not the power to do but to attract. As a result, women’s striving to attain it makes them feel inferior to what they actually are. They can not choose this power freely on her capacity is always under social censorship. That’s why the author objects women’s striving to attain beauty
What change in attitude do you think sontag wants to bring about in her female readers?
Sontag wants to bring a complete change in attitude in her female readers. The conventional attitude on beauty has confined women’s potentiality. The author wants to emphasize inner beauty- vision and wisdom, which is long lasting and more fruitfull. Woman should not limit their to be beautiful outwardly, if they feel that to be or to try to be beautiful is their aim in life. It will certainly make them more inferior and dependant to male. Without being beautifuk too women can act as complemently as males. She has been able to generate sympathy for women from them.
What do you think sontag is saying to beautiful women? How do you think they would respond?
Sontag is advocating for the inner beauty which is considered to be more important than outward beauty especially for women. Our Society always teaches women to be fair and beautiful. SO, most women think that their only aim and responsibility is to be beautiful. Such a concept has not only degraded women’s dignity but has also made them inferior and dependent. The author is telling these women to think over the uselessness of outer looks. Their beauty should not be used to attract men. They should be excellent, competant, independent thoughtfull in order to revive the ideal valve of beauty and preserve their identity in the society. WOmen should not be flattered by men.
The conventionally used beauty signifiesthe attraction of physical appearance. Beauty has been distinguished as inner beauty(character intellect and vision) and outer beauty(facial attraction, physique or sense of proportion of the body) which is attributed to women only. The author seeks to discredit this outer beauty because it has devalued the concept of beauty which signified the wholeness, the excellence and the virtue of human during the classical time.
Sontag wants to revive the old Greek concept of inner beauty. Soctates diciples had observed the inner beauty in their teacher although he was ugly looking person.
If beauty is a source of power, why does sontag object to wone’s striving to attain it?
Beauty is a form of power. She always looked at with suspicious eye even if she has good rise in work, politics, law, medicine, business or whatever. She is always pressure to confess that she stills works at being attractive. This power is not the power to do but to attract. As a result, women’s striving to attain it makes them feel inferior to what they actually are. They can not choose this power freely on her capacity is always under social censorship. That’s why the author objects women’s striving to attain beauty
What change in attitude do you think sontag wants to bring about in her female readers?
Sontag wants to bring a complete change in attitude in her female readers. The conventional attitude on beauty has confined women’s potentiality. The author wants to emphasize inner beauty- vision and wisdom, which is long lasting and more fruitfull. Woman should not limit their to be beautiful outwardly, if they feel that to be or to try to be beautiful is their aim in life. It will certainly make them more inferior and dependant to male. Without being beautifuk too women can act as complemently as males. She has been able to generate sympathy for women from them.
What do you think sontag is saying to beautiful women? How do you think they would respond?
Sontag is advocating for the inner beauty which is considered to be more important than outward beauty especially for women. Our Society always teaches women to be fair and beautiful. SO, most women think that their only aim and responsibility is to be beautiful. Such a concept has not only degraded women’s dignity but has also made them inferior and dependent. The author is telling these women to think over the uselessness of outer looks. Their beauty should not be used to attract men. They should be excellent, competant, independent thoughtfull in order to revive the ideal valve of beauty and preserve their identity in the society. WOmen should not be flattered by men.
Friday, May 28, 2010
Computer hierarchy
Supercomputers
The term supercomputer does not refer to a specific technology, nut to the fastest computing engines available at any give time. Super computers typically run military and scientific applications, although their use of commercial applications.
Mainframe
Mainframe remains popular in large enterprise for extensive computing applications that are accessed by thousands of users. Mainframe are less powerful and generally less expensive than supercomputer.
Minicomputers
are relatively small inexpensive and compact computers that perform the same functions as mainframe computers, but to a more limited extent. These computer are designed to accomplist specific taskes such as process control, scientific research and engineering applications
Microcomputer
Are the smallest and least expensive category of general purpose computers They can be sub divided into four classifications based on their size desktop thin clients notebooks and laptops.
The term supercomputer does not refer to a specific technology, nut to the fastest computing engines available at any give time. Super computers typically run military and scientific applications, although their use of commercial applications.
Mainframe
Mainframe remains popular in large enterprise for extensive computing applications that are accessed by thousands of users. Mainframe are less powerful and generally less expensive than supercomputer.
Minicomputers
are relatively small inexpensive and compact computers that perform the same functions as mainframe computers, but to a more limited extent. These computer are designed to accomplist specific taskes such as process control, scientific research and engineering applications
Microcomputer
Are the smallest and least expensive category of general purpose computers They can be sub divided into four classifications based on their size desktop thin clients notebooks and laptops.
Carrer in IT
Programmer
Programmer are IS professionals who modify existing computer programs or write new programs to satisfy user requirments,
Systems Analyst/ Developers
System analyst are information systems professionals who specialize in analyzinf and designing information systems.
System Operations Specialist
System operations specialist keep systems up and running and many have subspecializations with particular types of computing hardware(eg. mainfrane)
Business Analyst
A business analyst has IT experience and an in-depth knowledge of the organization's business processes. He or she is ofter heavily involved with the development of new information systems and acts as a translator between IS developers and users.
Database Administrator
Data are ab extremely valuable organizational asset for both day to day operations and strategic objectives. Database administrator typically have considerable experience and training in one or more types of database software and hardware.
Programmer are IS professionals who modify existing computer programs or write new programs to satisfy user requirments,
Systems Analyst/ Developers
System analyst are information systems professionals who specialize in analyzinf and designing information systems.
System Operations Specialist
System operations specialist keep systems up and running and many have subspecializations with particular types of computing hardware(eg. mainfrane)
Business Analyst
A business analyst has IT experience and an in-depth knowledge of the organization's business processes. He or she is ofter heavily involved with the development of new information systems and acts as a translator between IS developers and users.
Database Administrator
Data are ab extremely valuable organizational asset for both day to day operations and strategic objectives. Database administrator typically have considerable experience and training in one or more types of database software and hardware.
Business in the information age
PRESSURES
1) Global competition for trade and for labor
Low labor costs make chineses firms attractive as partners in joint manufacturing ventures
2)Changing Work force
An intresting number of females, single parents, minorities, and physically challenged persons work today in all types of postions. More employees than ever before prefer to defer retirement. It is easing the integration of these various employees inot the traditional workforce.
3)Customer orientation
Customer sophistication and expectations increses as customers become more knowledgeable about the availability and quality of products and services. Customers are demanding ever-more detailed information about products and services. They want to know what feature warranties are available.
4)Information overload
The Internet and other telecommunications nteworks increase the amount of information available to organization and individuals. The amount information available on the Internet more that doubles every year and most of it is free.
5)Social responsibility
Social issues affecting organizations range from the state of the plysical environment. to non discriminatory employment practices, to the spread of infectioud diseases and other health concerns. Failure to accept social responsibility can result in employee dissatisfaction and trunover, a tarnished corporate reputation with the public, and in some cases, governmental samctions.
ORGANIZATIONAL RESPONSES
1) Strategic systems
Organizations seek to implement systems that will significantly impact the organization's operations, sucess or survival. Such strtegic systems provide organizations with strange advantages in meeting organizational objectives, enabling them to increase their market share, to better negotiate with their suppliers, or to prevent competitord from entering their markets. There are a variety of IT-supported strategic systems.
2)Customer focus and service
Organization must may more attentions to customers and their preferences. Sometimes such an approach even involves reengineering the organization to better meet customer demands. This can be done in part by changing manufacturing process from mass production to mass customization. In mass production, a company produces a large quantity of titems that are manufactured to fit the desires of each customer. Information technology supports mass customization.
3)Continuous improvement efforts
In response to business pressures, many firms also make continous efforts to improve their productivity and quality.
4)Business alliances
In response to some of the competitive pressures of the global economy, many firms are realizing that alliances with other firms, even competitors, can be very beneficial.
1) Global competition for trade and for labor
Low labor costs make chineses firms attractive as partners in joint manufacturing ventures
2)Changing Work force
An intresting number of females, single parents, minorities, and physically challenged persons work today in all types of postions. More employees than ever before prefer to defer retirement. It is easing the integration of these various employees inot the traditional workforce.
3)Customer orientation
Customer sophistication and expectations increses as customers become more knowledgeable about the availability and quality of products and services. Customers are demanding ever-more detailed information about products and services. They want to know what feature warranties are available.
4)Information overload
The Internet and other telecommunications nteworks increase the amount of information available to organization and individuals. The amount information available on the Internet more that doubles every year and most of it is free.
5)Social responsibility
Social issues affecting organizations range from the state of the plysical environment. to non discriminatory employment practices, to the spread of infectioud diseases and other health concerns. Failure to accept social responsibility can result in employee dissatisfaction and trunover, a tarnished corporate reputation with the public, and in some cases, governmental samctions.
ORGANIZATIONAL RESPONSES
1) Strategic systems
Organizations seek to implement systems that will significantly impact the organization's operations, sucess or survival. Such strtegic systems provide organizations with strange advantages in meeting organizational objectives, enabling them to increase their market share, to better negotiate with their suppliers, or to prevent competitord from entering their markets. There are a variety of IT-supported strategic systems.
2)Customer focus and service
Organization must may more attentions to customers and their preferences. Sometimes such an approach even involves reengineering the organization to better meet customer demands. This can be done in part by changing manufacturing process from mass production to mass customization. In mass production, a company produces a large quantity of titems that are manufactured to fit the desires of each customer. Information technology supports mass customization.
3)Continuous improvement efforts
In response to business pressures, many firms also make continous efforts to improve their productivity and quality.
4)Business alliances
In response to some of the competitive pressures of the global economy, many firms are realizing that alliances with other firms, even competitors, can be very beneficial.
Problems with Tradional Data base approach
Data Redundancy
As applications and their data files were created by different programmer over a period of time, the same piece of information could be duplicated in several places. Resulting in overconsuming of space
Data inconsistency:
Means that the various copies of the data no longer agree.
Data Isolation:
It also leads ro difficulty in accessing data from different applicatiob
Security:
is difficult to enforce in the file environment beacuse new applications may be added to the system on an ad-hoc basis and with more applications more people have access to data
As applications and their data files were created by different programmer over a period of time, the same piece of information could be duplicated in several places. Resulting in overconsuming of space
Data inconsistency:
Means that the various copies of the data no longer agree.
Data Isolation:
It also leads ro difficulty in accessing data from different applicatiob
Security:
is difficult to enforce in the file environment beacuse new applications may be added to the system on an ad-hoc basis and with more applications more people have access to data
Programming language
Machine Language
is the lowest level computer language, consisting of the internal representation of instructions and data. This machine code-the actual instructions understood an directly executablr by the central processing unit-- is composed of binary digits.
Assembly Language
is the next level up from machine language. It is still considered a lower level language but is more user friendly because it represents machine language instructions and data locatioins in promary storage.
Procedural Language
are the next step in the evolution of user-oriented programming languages. They are also called third-generation languages or 3GLs. Procedural languages are much closer to so-called natural language and therefore are easier to write read and alter
Nonprocedural Language
allows the user to specify the desired result without having to specify the detailed procedures needed for achiveing the result. These languages are fourth-generation languages.
Natural Programming Languages
NPL are the next evolutionary steps. They are sometimes know as fifth generation languages. nMost of these language are still experimental and have yet to be widely adopted by industry
is the lowest level computer language, consisting of the internal representation of instructions and data. This machine code-the actual instructions understood an directly executablr by the central processing unit-- is composed of binary digits.
Assembly Language
is the next level up from machine language. It is still considered a lower level language but is more user friendly because it represents machine language instructions and data locatioins in promary storage.
Procedural Language
are the next step in the evolution of user-oriented programming languages. They are also called third-generation languages or 3GLs. Procedural languages are much closer to so-called natural language and therefore are easier to write read and alter
Nonprocedural Language
allows the user to specify the desired result without having to specify the detailed procedures needed for achiveing the result. These languages are fourth-generation languages.
Natural Programming Languages
NPL are the next evolutionary steps. They are sometimes know as fifth generation languages. nMost of these language are still experimental and have yet to be widely adopted by industry
Computer Memory
Primary Storage
or main memory as it is sometimes called stores for very brief periods of time three types of information: data to be processed ny the CPU, instructions for the CPU, as how to process the data and operating system programs that manager various aspects of the computers's operation.
Registers.
Registers are part of the cpu . They have the least capacity, storing extremely limited amounts of instructions and data only immediately before and after processing.
Random access memory
It stores more information than the registers and is farther away from the CPU, but it stores less than secondary storage and is much closer to the CPU than is Secondary storage. RAM is temporary and volatile
Cache memory
It is a type of high speed memory that a processor can access more rapidly than main memory. Cache memory is a place closer to the CPU where the computer can temporarily store blocks of instructions used most often.
Read- only memory
is the place( atype of chip) where certain critical tinstructions are safeguarded. ROM is not volatile and retain these instructions when the power to the computertruned off. The read-only designation means that these instruction can be read only by the computer and can not be changed by user.
SECONDARY STORAGE
Magnetic media
Magnetic diskettes
Optical Storage
Memory cards
or main memory as it is sometimes called stores for very brief periods of time three types of information: data to be processed ny the CPU, instructions for the CPU, as how to process the data and operating system programs that manager various aspects of the computers's operation.
Registers.
Registers are part of the cpu . They have the least capacity, storing extremely limited amounts of instructions and data only immediately before and after processing.
Random access memory
It stores more information than the registers and is farther away from the CPU, but it stores less than secondary storage and is much closer to the CPU than is Secondary storage. RAM is temporary and volatile
Cache memory
It is a type of high speed memory that a processor can access more rapidly than main memory. Cache memory is a place closer to the CPU where the computer can temporarily store blocks of instructions used most often.
Read- only memory
is the place( atype of chip) where certain critical tinstructions are safeguarded. ROM is not volatile and retain these instructions when the power to the computertruned off. The read-only designation means that these instruction can be read only by the computer and can not be changed by user.
SECONDARY STORAGE
Magnetic media
Magnetic diskettes
Optical Storage
Memory cards
It support at diffferent organizational levels
Strategic decisions
are usually made by top managment these are relatively long term planning decisions that deal with the organization's objectives as a whole and the allocation of resources to achive these objectives.
Tactical or managerial decisions
are made by middle managers, who prepare short-term plans, procedures, and policies with which to begin implementing the organization's long term strategies. MISs provide the primary support at this level, along with some types of DSS
Operational decisions
are made by line managers and operators. These are the day-to-day decisions that aim to keep the organization's operations moving smoothly. TPSs typically capture the operational information relevant for decision making this level.
Pyramin diagram
Strategic System-----------Top Managers
Staff Support------------------------Knowledge workers, Professionals
MAnagerial systems--------------------------------Middle Managers
Operational System-------------------------------------------Line managars, operators
Office automation and Communication system-----------------------Clerical Staff
are usually made by top managment these are relatively long term planning decisions that deal with the organization's objectives as a whole and the allocation of resources to achive these objectives.
Tactical or managerial decisions
are made by middle managers, who prepare short-term plans, procedures, and policies with which to begin implementing the organization's long term strategies. MISs provide the primary support at this level, along with some types of DSS
Operational decisions
are made by line managers and operators. These are the day-to-day decisions that aim to keep the organization's operations moving smoothly. TPSs typically capture the operational information relevant for decision making this level.
Pyramin diagram
Strategic System-----------Top Managers
Staff Support------------------------Knowledge workers, Professionals
MAnagerial systems--------------------------------Middle Managers
Operational System-------------------------------------------Line managars, operators
Office automation and Communication system-----------------------Clerical Staff
Information System
Information system collects, processes, stores, analyzes, and disseminates information for a specific purpose. Like any other system, an information system includes inputs(data, instructions) and outputs(reports, calculations). It processes the inputs and produces outputs that are sent to the users or other system.
Data: are raw facts or elementary descriptions of things, events, activities, and transactions that are captured, recorded, stored and classified but not organized to convey any specific meaning. example grade point average, bank balance
Information: it is a collection of facts(data) organized in some manner so that they are meaningful to recipient. example student name with grade point average, customer name with bank balance.
Knowledge: consist of information that has been organized and processed to convey understanding, experiences, accumulated learning or expertise as it applies to a current business problem or process.
A computer based information system is an information system that used computer and telecommunication s technology to perform its intended task
Components of an Computer based information system
The compinents of an information system are hardware, software, database, net-works, procedures and people.
Hardware is a set of device sucs as a processor, monitor, keyboard and printer that accept data and information, process them and display them.
Software is a set of computer program that enable the hardware to process data.
A database is an organized collection of related files and records that stores data and the association among them.
A networks is a connecting system that permits the sharing of resources among different computers.
Procedures are the strategies, policies, method and rules for using the information system
The most important element in information systems is people, those persons who work with the information system or use its output
Cababilities of information system
1) Provide fast and accurate transaction processing
Every event that occurs in a business is called a transaction. Transactions include the sale of a unit of goods, a pay-check issued, a bank deposit, a course grade registered and so on. These data nust be captured accurately and quickly. This process is called transaction processing. A good example of transaction processing system ins point of sale.
2)Provide large-capacity, fast-access storage
Information systems must provide both enormous storage for corporate data, and also fast access to those data.
3)Provide fast communications(machine to machine, human to human)
Networks enable organozational employees and computers to communicate almost instantly around the world. High-transmission capacity networks(those with high bandwidths) make fast communications possible. In addition they allow data, voice, image, documents, and full-motion video to be transmitted simultaneously
4)Reduce information overload
Information systems(particularly networks) have contributed to managers having too much information. For example, the amount of sult, managers can feel drowned in information and unable to make decisions efficiently and effectively. Information systems can be designed to reduce this information overload.
5)Provide support for decision making
Decision support systems help decision makers across an organization and at all levels of the organization. Therefore, employees at lower organizationals levels have the authority and responsibility to make more and larger decisions than ever before.
Strategic use of information system
The forces that shape the level and type of competition in any industry include the relative power of buyers and suppliers, threats from substitute products and services, and the ease or difficulty with which new competitors can enter the industry. BAsic competitive strategies emphasize cost reduction and product differentiation. In addition, the value chain enables companies to understand and optimize the discrete steps involved in the manufacture of a product or the provision of a service, so as to add value to their products by improving quality and efficiency at every steps. These business concepts can be used to generate a series of questions about IT whose answers can shape an organization's competitive strategy. In general, IT give strategic competitive advantage through addressing and sometimes altering the nature of strategic competitive advantage through addressong and sometimes altering the nature of strategic force in industry, as well as advancing strategies based on cost and product differentiation
Major Is function
# Initiating and designing specific strategic information systems
# Infrastructure planning, development, and control
# Incorporating the Internet and e-commerce into the business
# Educating the non-IS managers about IT
# Supporting end-user computing
# Partnering with the executive level that runs the businerr
Data: are raw facts or elementary descriptions of things, events, activities, and transactions that are captured, recorded, stored and classified but not organized to convey any specific meaning. example grade point average, bank balance
Information: it is a collection of facts(data) organized in some manner so that they are meaningful to recipient. example student name with grade point average, customer name with bank balance.
Knowledge: consist of information that has been organized and processed to convey understanding, experiences, accumulated learning or expertise as it applies to a current business problem or process.
A computer based information system is an information system that used computer and telecommunication s technology to perform its intended task
Components of an Computer based information system
The compinents of an information system are hardware, software, database, net-works, procedures and people.
Hardware is a set of device sucs as a processor, monitor, keyboard and printer that accept data and information, process them and display them.
Software is a set of computer program that enable the hardware to process data.
A database is an organized collection of related files and records that stores data and the association among them.
A networks is a connecting system that permits the sharing of resources among different computers.
Procedures are the strategies, policies, method and rules for using the information system
The most important element in information systems is people, those persons who work with the information system or use its output
Cababilities of information system
1) Provide fast and accurate transaction processing
Every event that occurs in a business is called a transaction. Transactions include the sale of a unit of goods, a pay-check issued, a bank deposit, a course grade registered and so on. These data nust be captured accurately and quickly. This process is called transaction processing. A good example of transaction processing system ins point of sale.
2)Provide large-capacity, fast-access storage
Information systems must provide both enormous storage for corporate data, and also fast access to those data.
3)Provide fast communications(machine to machine, human to human)
Networks enable organozational employees and computers to communicate almost instantly around the world. High-transmission capacity networks(those with high bandwidths) make fast communications possible. In addition they allow data, voice, image, documents, and full-motion video to be transmitted simultaneously
4)Reduce information overload
Information systems(particularly networks) have contributed to managers having too much information. For example, the amount of sult, managers can feel drowned in information and unable to make decisions efficiently and effectively. Information systems can be designed to reduce this information overload.
5)Provide support for decision making
Decision support systems help decision makers across an organization and at all levels of the organization. Therefore, employees at lower organizationals levels have the authority and responsibility to make more and larger decisions than ever before.
Strategic use of information system
The forces that shape the level and type of competition in any industry include the relative power of buyers and suppliers, threats from substitute products and services, and the ease or difficulty with which new competitors can enter the industry. BAsic competitive strategies emphasize cost reduction and product differentiation. In addition, the value chain enables companies to understand and optimize the discrete steps involved in the manufacture of a product or the provision of a service, so as to add value to their products by improving quality and efficiency at every steps. These business concepts can be used to generate a series of questions about IT whose answers can shape an organization's competitive strategy. In general, IT give strategic competitive advantage through addressing and sometimes altering the nature of strategic competitive advantage through addressong and sometimes altering the nature of strategic force in industry, as well as advancing strategies based on cost and product differentiation
Major Is function
# Initiating and designing specific strategic information systems
# Infrastructure planning, development, and control
# Incorporating the Internet and e-commerce into the business
# Educating the non-IS managers about IT
# Supporting end-user computing
# Partnering with the executive level that runs the businerr
Wednesday, May 26, 2010
Interference:
When teo light easves travels in a same region, then there will be change in intensity due to superposition of light waves. This phenomenon is called interference. In other world non uniform distribution of energy when two light source super impose.
COndition for maxima and minima
Derive an expression for intensity on different points
let us consider two light waves travelling in a same region
let, y1 = asinwt ------1
y2 = ain(wt + theta)--------2
be the equation of two waves. We considered same amplitude, angular velocity but slightly different phase.
Here phase difference = theta
Now, the resultant wave can be given as
y = y1 + y2
= asinwt + asin(wt + theta)
=a(sinwt + sin(wt + theta))
y=(2acos(theta/2)*(sinwt+(theta/2))-------3
here, A = 2acos(theta/2)
We know intensity is directly proportional to A^2
therefor I directly porportional to A^2
I = A^2
I = 4a^2(cos(theta/2))
for maxima,
cos^2(theta/2) = 1
cos(theta/2)=+-1
cos(theta/2) = cos(n(pie)), n = 0,1,2....
or theta/2 = n(pie)
therefore theta = 2npie
here theta phase difference = lamda/2pie * theta phase difference = n*lamda
for minima cos^2(theta/2) = 0
or cos(theta/2) = 0
cos(theta/2) = cos(2n+1/2)pie
theta = (2n+1)pie
corresponding path difference
= lamda/2pie *(2n + 1)/2*pie
=(2n+1)/2 * lamda
maximum point 0, lamda,2lamda,3lamda
minimum points lamda/2 ,3/2lamda 5/2lamda
COndition for maxima and minima
Derive an expression for intensity on different points
let us consider two light waves travelling in a same region
let, y1 = asinwt ------1
y2 = ain(wt + theta)--------2
be the equation of two waves. We considered same amplitude, angular velocity but slightly different phase.
Here phase difference = theta
Now, the resultant wave can be given as
y = y1 + y2
= asinwt + asin(wt + theta)
=a(sinwt + sin(wt + theta))
y=(2acos(theta/2)*(sinwt+(theta/2))-------3
here, A = 2acos(theta/2)
We know intensity is directly proportional to A^2
therefor I directly porportional to A^2
I = A^2
I = 4a^2(cos(theta/2))
for maxima,
cos^2(theta/2) = 1
cos(theta/2)=+-1
cos(theta/2) = cos(n(pie)), n = 0,1,2....
or theta/2 = n(pie)
therefore theta = 2npie
here theta phase difference = lamda/2pie * theta phase difference = n*lamda
for minima cos^2(theta/2) = 0
or cos(theta/2) = 0
cos(theta/2) = cos(2n+1/2)pie
theta = (2n+1)pie
corresponding path difference
= lamda/2pie *(2n + 1)/2*pie
=(2n+1)/2 * lamda
maximum point 0, lamda,2lamda,3lamda
minimum points lamda/2 ,3/2lamda 5/2lamda
Young's double slit experiment
Let us consider a source of monochromatic light s. The light from s is passed through two slits s1 and s2. So that s1 and s2 behaves as two coherent source according to hygen's principal. let seperation between s1 adn s2 be d. COnsider a screen be place at distance D from two slits. Now when light from s1 and s2 reach the screen interference effect will result. COnsider a point p on the screen at distance y from center of screen. Now the path difference between lights from s1 and s2 at point P can be calculated which is S2N in fig.
Join C&P and suppose < PCO = theta
Let s1, s2 and N are very close to eath other, so S2N is approximately perpendicular to CP
Then,
sintheta = s2N/S1S2
s2N = dsintheta ---------1
Sp, point P will be bright fringe if
S2N = dsintheta = nlamda
or sintheta = nlamda/d ----------2
Now in triangle CPO, tantheta = PO/CO = y/D
Here in actual practice, theta is very small and for very small theta, sintheta = tantheta
y/D = nlamda/d
y = nlamdaD/d, n =0,1,2,3
for n = 0 n = 1 n = 2
y0 = 0 y1 = lamdaD/d y2 = 2lamdaD/d
Distance between two consecutive bright fringe = lamdaD/d
point p will be dark fringe if,
S2N = dsintheta = (2n+1)/2 * lamda
sintheta = (2n+1)/2 * lamda/d ----------4
From equation 3 and 4 for very small theta (sintheta = tantheta)
y/D = (2n+1)/2 * lamda/d
or y = (2n+1)/2 * lamda*D /d----------------5
Now, for
n=0 n=1
y0 = lamda*D/2d y1 = 3/2 lamdaD/d y2 = 5/2*lamdaD/d
fringe width = beta*D/d
Fringe width: The distance between two consequtive bright or dark fringe is called fringe width and denoted by beta
beta = lamda*D/d
to calculate lamda,
beta = lamda * D/d
lamda = beta*d/D
Join C&P and suppose < PCO = theta
Let s1, s2 and N are very close to eath other, so S2N is approximately perpendicular to CP
Then,
sintheta = s2N/S1S2
s2N = dsintheta ---------1
Sp, point P will be bright fringe if
S2N = dsintheta = nlamda
or sintheta = nlamda/d ----------2
Now in triangle CPO, tantheta = PO/CO = y/D
Here in actual practice, theta is very small and for very small theta, sintheta = tantheta
y/D = nlamda/d
y = nlamdaD/d, n =0,1,2,3
for n = 0 n = 1 n = 2
y0 = 0 y1 = lamdaD/d y2 = 2lamdaD/d
Distance between two consecutive bright fringe = lamdaD/d
point p will be dark fringe if,
S2N = dsintheta = (2n+1)/2 * lamda
sintheta = (2n+1)/2 * lamda/d ----------4
From equation 3 and 4 for very small theta (sintheta = tantheta)
y/D = (2n+1)/2 * lamda/d
or y = (2n+1)/2 * lamda*D /d----------------5
Now, for
n=0 n=1
y0 = lamda*D/2d y1 = 3/2 lamdaD/d y2 = 5/2*lamdaD/d
fringe width = beta*D/d
Fringe width: The distance between two consequtive bright or dark fringe is called fringe width and denoted by beta
beta = lamda*D/d
to calculate lamda,
beta = lamda * D/d
lamda = beta*d/D
Newton's ring
Let s be a source of monochromatic light. light from s are made parallel be a lense L which are then allowed to fall on a glass plate inclined at 45degree with horizontal part of light will be trasnmitted and some part will be reflected down. The reflected lights then allowed to fall on the plane surface of a plano convex lense place on a base plate with convex surface touching base plate. Now, some part of light will be reflected from the lower(convex) surface of plano-convex lens and some parts will be reflected from upper surface of base plate. During this process, there will be interference in reflected lights because there is a thin film of air introduced in between plano-convex lense and base plate. The fringes will be circular and called as Newton's ring.
Here interference is because of thin film of air So,
for bright fringr
2mew t cosr = (2n+1)/2 * lamda-------------1
n = 0,1,2
For dark fringe
2 mew t cos r = n*lamda -------------------2
n=0,1,2
Diagram
Let a ring be observed at height(thickness) OD = t at a distance r from the point of contact.
Now, from fig. we can have
AD*DB = ED*OD
r*r = (EO-OD)*OD
or r^2 = (2R - t)* t
As t is very small, we can have
r^2 = 2Ry
t = r^2/2R
Now, condition for bright fringe will be
2mew * r^2/2R * cosr = (2n+1)/2 * lamda
if r is very small, cosr = 1
or r^2 = 2n+1/2* R*lamda/mew
r =???? --------------4mew = 1
n= 0,1,2.....
For DArk fringe
2mew * r^2/2R = n*lamda
r=??????
n = 0,1,2......
Here interference is because of thin film of air So,
for bright fringr
2mew t cosr = (2n+1)/2 * lamda-------------1
n = 0,1,2
For dark fringe
2 mew t cos r = n*lamda -------------------2
n=0,1,2
Diagram
Let a ring be observed at height(thickness) OD = t at a distance r from the point of contact.
Now, from fig. we can have
AD*DB = ED*OD
r*r = (EO-OD)*OD
or r^2 = (2R - t)* t
As t is very small, we can have
r^2 = 2Ry
t = r^2/2R
Now, condition for bright fringe will be
2mew * r^2/2R * cosr = (2n+1)/2 * lamda
if r is very small, cosr = 1
or r^2 = 2n+1/2* R*lamda/mew
r =???? --------------4mew = 1
n= 0,1,2.....
For DArk fringe
2mew * r^2/2R = n*lamda
r=??????
n = 0,1,2......
Hygens Principal Geometrical Path
Wave front: Locus of thr particle having same phase of medium in which a wave travel is called wavefront and a perpendicular drawn to wavefront away from source is called ray of light
Hygen’s Principal: Hygens’s principle states that
1. every point on the primary wavefront behaves as the source of secondary wavefront in all possible direction
2. The direction of secondary wavefront is given by the forward envelop on the primary wavefront
Diagram of a circle with small circle in its circumference
Let us consider s be the source of light which emit light waves in all possible direction. At time t light will move distance ct and forms a primary wavefront of radius ct. The point like a,b,c,d.... on primary wavefront are now the source of secondary wavefront and emit light in all possible direction. Again in further time delta t the source like a,b,c,d... forms the wave front of radius c delta t. The locus touching all the tangent points in forward direction in wavefront of source a,b,c,d... is the direction of secondary wavefront
Geometrical and optical path
The path travelled by the light in air or vaccum is called the optical path and the path travelled by light in certain medium is called geometrical path
Diagram with a arrow passing through a rectangle called medium
Let us consider light travel a distance x in certain medium maving refractive index meu in time t then
Geometrical path x = vt
where v = velocity of light in medium
t = x/v
Now we can have the optical path in same time
optical path = ct = c*x/v = c/v*x = meu *x = refractive index * geometrical path
Hygen’s Principal: Hygens’s principle states that
1. every point on the primary wavefront behaves as the source of secondary wavefront in all possible direction
2. The direction of secondary wavefront is given by the forward envelop on the primary wavefront
Diagram of a circle with small circle in its circumference
Let us consider s be the source of light which emit light waves in all possible direction. At time t light will move distance ct and forms a primary wavefront of radius ct. The point like a,b,c,d.... on primary wavefront are now the source of secondary wavefront and emit light in all possible direction. Again in further time delta t the source like a,b,c,d... forms the wave front of radius c delta t. The locus touching all the tangent points in forward direction in wavefront of source a,b,c,d... is the direction of secondary wavefront
Geometrical and optical path
The path travelled by the light in air or vaccum is called the optical path and the path travelled by light in certain medium is called geometrical path
Diagram with a arrow passing through a rectangle called medium
Let us consider light travel a distance x in certain medium maving refractive index meu in time t then
Geometrical path x = vt
where v = velocity of light in medium
t = x/v
Now we can have the optical path in same time
optical path = ct = c*x/v = c/v*x = meu *x = refractive index * geometrical path
Diffraction
The bending of light from the corners of bostacle placed in path and spreading of light to the geometrical shadow is called comparable with the wavelength of light used then diffracted rays will overlay and results the diffraction fringe called as secondary minima and secondary maxima
Types od Diffraction
1) Freshel's diffraction: When diffraction is observed placing source or screen or both of them at a finite distance from slit. Then this is called as Fresnel's diffraction
2)Frauhafer diffraction
when source and screen are placed at infinte distance from slit and diffraction is observed then this types of diffraction is called fraunhofer diffraction. Lenses are used to observe this type of diffraction
Diffraction through Single slit
Let a parallel beam of light be incident on a slit of widtha then beam will be diffracted and we can observe a diffraction pattern on the screen, When slit width a is comparable with wavelength diffraction fringes can be observed.
Principal maxima: At the centre point of screen all the wave from slit will reach is same phase i.e path difference will be zero and hence intensity will be maximum or bright fringe occurs and called as principla maxima
Secondary minima
Let us consider lamda be the wavelength of the light used and a be the width of the slit. Now, divide the slit into two equal halves. Then taking waves travelling at an angle theta towards screen. Then path difference between 1st and 2nd wave will be
S2N = S1S1sintheta
= a/2 * sintheta
path difference = a/2 sin theta
If this path difference is ewusl to lamda/2 then dark fringe occurs.
i.e a*2 * sin theta = lamda/2
asin theta = lamda---------------1
Now if the slit is divided into four equal parts then path difference will be qual parts, then path difference will be equal to a/4 sin theta and dark fringe would occur if
a/4 * sin theta = lamda/2
a= sintheta = 2 lamda ------------2
and so on
asin theta = n lamda-----------------3
n = 1,2,3...................
FOr first secondary minima
asin theta = lamda
2nd secondary minima, asin theta = 2 lamda
Secondary maxima
When the slit is divide into three equal parts and two consecutive wave calcels effect6s of eath other and third remains unaffrected and gives bright fringe i.e
For first secondary maxima,
asin theta = 3/2 * lamda
2nd secondary maxima
asin theta = (2n+1)/2 * lamda
n= 1,2.........
Diffraction through double slit
Let us consider AB and CD be two identical slits having width a separated by an opaque BC of width b. Let a beam of monochromatic light be incident on the slits then interference and diffraction pattern can be observed on the screen.
Interference maxima and minima
At the centre of screen all the waves will meet in same phase so, this point will be of maximum intensity and called as principal maxima.
DIagram
Every point on the slit behaves as secondary wave source. Consider the central points of eath slits as source such that they will be coherent. If we take wave moving at anglr theta towards screen then path difference between these two S2N.
S2N = S1S2sintheta
=(a+b)sintheta
Now, interference maxima would occur if (a+b)sin theta = n lamda and interference minima would occur if
(a+b)sin theta = (2n+1)/2 * lamda
Diffraction maxima and minima
Let us consider two waves diffrented from upper edge of each slit moving at an angle theta towards screen. Then path difference between these two will be
S2N = S1S2 sin theta
=(a+b) sin theta
Diffraction maxima would occur if
(a+b) sin theta = n * lamda
and Diffraction minima would occur if
(a+b)sin theta = (2n+1)/2 * lamda
Diffraction grating
It is an arrangment of larger number of ruling ina glass plate of 1 inch in which the ruled lines behaves as opaque portion and glass left between each two line behaves as slit. Generally there are around 2000 lines in a glass plate of 1 inch.
If ruling is made on a transparent glass plate, the it is called a plain transmission grating and if it is made on a mirror then it is called as reflection grating. Here, combination of a single slit and a single opaque is called as grating element
If N be the no. of lines in a glass plate of one inch and a and b be the width of slit and opaque portion then
Na+Nb = 1 inch
(a+b) = 1/N inch = 2.54/N cm
Types od Diffraction
1) Freshel's diffraction: When diffraction is observed placing source or screen or both of them at a finite distance from slit. Then this is called as Fresnel's diffraction
2)Frauhafer diffraction
when source and screen are placed at infinte distance from slit and diffraction is observed then this types of diffraction is called fraunhofer diffraction. Lenses are used to observe this type of diffraction
Diffraction through Single slit
Let a parallel beam of light be incident on a slit of widtha then beam will be diffracted and we can observe a diffraction pattern on the screen, When slit width a is comparable with wavelength diffraction fringes can be observed.
Principal maxima: At the centre point of screen all the wave from slit will reach is same phase i.e path difference will be zero and hence intensity will be maximum or bright fringe occurs and called as principla maxima
Secondary minima
Let us consider lamda be the wavelength of the light used and a be the width of the slit. Now, divide the slit into two equal halves. Then taking waves travelling at an angle theta towards screen. Then path difference between 1st and 2nd wave will be
S2N = S1S1sintheta
= a/2 * sintheta
path difference = a/2 sin theta
If this path difference is ewusl to lamda/2 then dark fringe occurs.
i.e a*2 * sin theta = lamda/2
asin theta = lamda---------------1
Now if the slit is divided into four equal parts then path difference will be qual parts, then path difference will be equal to a/4 sin theta and dark fringe would occur if
a/4 * sin theta = lamda/2
a= sintheta = 2 lamda ------------2
and so on
asin theta = n lamda-----------------3
n = 1,2,3...................
FOr first secondary minima
asin theta = lamda
2nd secondary minima, asin theta = 2 lamda
Secondary maxima
When the slit is divide into three equal parts and two consecutive wave calcels effect6s of eath other and third remains unaffrected and gives bright fringe i.e
For first secondary maxima,
asin theta = 3/2 * lamda
2nd secondary maxima
asin theta = (2n+1)/2 * lamda
n= 1,2.........
Diffraction through double slit
Let us consider AB and CD be two identical slits having width a separated by an opaque BC of width b. Let a beam of monochromatic light be incident on the slits then interference and diffraction pattern can be observed on the screen.
Interference maxima and minima
At the centre of screen all the waves will meet in same phase so, this point will be of maximum intensity and called as principal maxima.
DIagram
Every point on the slit behaves as secondary wave source. Consider the central points of eath slits as source such that they will be coherent. If we take wave moving at anglr theta towards screen then path difference between these two S2N.
S2N = S1S2sintheta
=(a+b)sintheta
Now, interference maxima would occur if (a+b)sin theta = n lamda and interference minima would occur if
(a+b)sin theta = (2n+1)/2 * lamda
Diffraction maxima and minima
Let us consider two waves diffrented from upper edge of each slit moving at an angle theta towards screen. Then path difference between these two will be
S2N = S1S2 sin theta
=(a+b) sin theta
Diffraction maxima would occur if
(a+b) sin theta = n * lamda
and Diffraction minima would occur if
(a+b)sin theta = (2n+1)/2 * lamda
Diffraction grating
It is an arrangment of larger number of ruling ina glass plate of 1 inch in which the ruled lines behaves as opaque portion and glass left between each two line behaves as slit. Generally there are around 2000 lines in a glass plate of 1 inch.
If ruling is made on a transparent glass plate, the it is called a plain transmission grating and if it is made on a mirror then it is called as reflection grating. Here, combination of a single slit and a single opaque is called as grating element
If N be the no. of lines in a glass plate of one inch and a and b be the width of slit and opaque portion then
Na+Nb = 1 inch
(a+b) = 1/N inch = 2.54/N cm
Tuesday, May 25, 2010
Monochromatic Abberebation
Monochromatic aberration
1) Spherical aberration
The focal length of a lense for marginal and para-axial rays varies because of the thickness of lense. It os maximus for paraxial and minimum for marginal rays. This inability of lense to focus all rays toa single point is called spherical aberration fp-fm is measure of spherical aberration
Removal
a) If two lenses are kept at a distance equal to difference in their focal length spherical aberration will be minimized
b)If two(convex and concave) lense are kept in a suitable combination spherical aberration can be removed
c)By using Stops for marginal or for para-axial.
2)Comatic aberration
Spreading of image in the direction perpendicular to principal axis is called coma. The inability of a lense to focus all the incident light on a vertical plane from the object placed outside the axis is called comatic aberration
Removal
a) By using stop chromatic aberration can be minimized
b) By using lense which obeys Abbe's sine law
3)Astigmatism
The inability of a lense to focus all the incident rays in vertical and horizontal plane results spreading of image parallel to principal axis is called astigmatism
Removal
1) By using stops
4) Distortion
If a rectangular object be place infront of a lens, image will be formed either in barrel shaped or pin-cushioned shape. This defect is called distortion.
Removal
1) using stops
2) using suitable combination of two conver lense
Curvature of field
A lens has variable focal length for marginal and paraxial rays. Due to this image will be formed in a curved shape and called as curvature of field
1) Spherical aberration
The focal length of a lense for marginal and para-axial rays varies because of the thickness of lense. It os maximus for paraxial and minimum for marginal rays. This inability of lense to focus all rays toa single point is called spherical aberration fp-fm is measure of spherical aberration
Removal
a) If two lenses are kept at a distance equal to difference in their focal length spherical aberration will be minimized
b)If two(convex and concave) lense are kept in a suitable combination spherical aberration can be removed
c)By using Stops for marginal or for para-axial.
2)Comatic aberration
Spreading of image in the direction perpendicular to principal axis is called coma. The inability of a lense to focus all the incident light on a vertical plane from the object placed outside the axis is called comatic aberration
Removal
a) By using stop chromatic aberration can be minimized
b) By using lense which obeys Abbe's sine law
3)Astigmatism
The inability of a lense to focus all the incident rays in vertical and horizontal plane results spreading of image parallel to principal axis is called astigmatism
Removal
1) By using stops
4) Distortion
If a rectangular object be place infront of a lens, image will be formed either in barrel shaped or pin-cushioned shape. This defect is called distortion.
Removal
1) using stops
2) using suitable combination of two conver lense
Curvature of field
A lens has variable focal length for marginal and paraxial rays. Due to this image will be formed in a curved shape and called as curvature of field
Optical Fibre
Optical Fibre
It is a glass fibre of diameter around 125um of high refractive index in the core coated by a glass with low refractive index. Now a ray of light incident on one end will emerge out through other end suffering multiple total internal reflection without losing energy.
Types of optical fibre
1) Monomode: Optical fibre in which inner glass rod is thinner as compared to outer coating
2)Multimode: Optical fibre in which inner glass is thicker as compared to outer coating
on the basis of refractive index of glass used
1)Step index: Optical fibre which consist of glass rod of high refractive index of 1.52 in core and glass of refractive index 1.48 in coating
2)Graded Index: Optical fibre which have glass of high refractive index goes on decreasing gradually from core to surface.
Advantage of Graded index
In case of step index when a wave enters into the glass rod suffers disperson and so a wave will be total internally reflected from different point as shown in figure so that they will come out from other end in different time. But in case of graded index the wave after dispersion will be totslly internally reflected from diffrent height as shows in figure. When they suffer total internal reflection on lower side they will converge again in single point at different place so that single can be obtained at a single time from other end.
The minumum angle of incidence in air which makes total internal reflection possible in glass is called acceptance angle.
We have for total internal reflection
((meu))=1/(sinC)==>sinC=1/((meu))
This implies
sinC = ((meu1))/((meu2)) ------1
let us consider an incident ray be incidenting in glass making angle of incidence in air i such that ((meu1)) >((meu2)), total internal reflection r = 90 - 2
((meu2)) = sini/sinr = sini/sin(90-c) = sini/cosC
cosC = sinI/((meu2))--------------2
squaring and adding 1 and 2 will give i=sin(inverse)*sqrt((meu1)square - (men2)square)) This is the value of minimum angle of acceptance
Use of optical fibre
1 It is used to transfer electronic data or signal
2 It is used in telecommunication
3 It is used for direction signalling
4 It is used in medical purpose as optical pipe
5 It is used in computer networking
Advantage if optical fibre
1 It is fast and we can transfer signals without loss of much energy
2 It is cheap to be used for long distance
3 Repitors can be kept at about 35 km whereas they must be kept at about 1 km in other cables
4 It is most secured
5 It has very large bandwidth from 10^13 hz--10^16Hz
Disadvantage of optical fibre
1 It is expensive for short distance
2 Highly skilled manpower is needed for its maintenance
It is a glass fibre of diameter around 125um of high refractive index in the core coated by a glass with low refractive index. Now a ray of light incident on one end will emerge out through other end suffering multiple total internal reflection without losing energy.
Types of optical fibre
1) Monomode: Optical fibre in which inner glass rod is thinner as compared to outer coating
2)Multimode: Optical fibre in which inner glass is thicker as compared to outer coating
on the basis of refractive index of glass used
1)Step index: Optical fibre which consist of glass rod of high refractive index of 1.52 in core and glass of refractive index 1.48 in coating
2)Graded Index: Optical fibre which have glass of high refractive index goes on decreasing gradually from core to surface.
Advantage of Graded index
In case of step index when a wave enters into the glass rod suffers disperson and so a wave will be total internally reflected from different point as shown in figure so that they will come out from other end in different time. But in case of graded index the wave after dispersion will be totslly internally reflected from diffrent height as shows in figure. When they suffer total internal reflection on lower side they will converge again in single point at different place so that single can be obtained at a single time from other end.
The minumum angle of incidence in air which makes total internal reflection possible in glass is called acceptance angle.
We have for total internal reflection
((meu))=1/(sinC)==>sinC=1/((meu))
This implies
sinC = ((meu1))/((meu2)) ------1
let us consider an incident ray be incidenting in glass making angle of incidence in air i such that ((meu1)) >((meu2)), total internal reflection r = 90 - 2
((meu2)) = sini/sinr = sini/sin(90-c) = sini/cosC
cosC = sinI/((meu2))--------------2
squaring and adding 1 and 2 will give i=sin(inverse)*sqrt((meu1)square - (men2)square)) This is the value of minimum angle of acceptance
Use of optical fibre
1 It is used to transfer electronic data or signal
2 It is used in telecommunication
3 It is used for direction signalling
4 It is used in medical purpose as optical pipe
5 It is used in computer networking
Advantage if optical fibre
1 It is fast and we can transfer signals without loss of much energy
2 It is cheap to be used for long distance
3 Repitors can be kept at about 35 km whereas they must be kept at about 1 km in other cables
4 It is most secured
5 It has very large bandwidth from 10^13 hz--10^16Hz
Disadvantage of optical fibre
1 It is expensive for short distance
2 Highly skilled manpower is needed for its maintenance
Cardinal Points
Focal Points
If a parallel beam of light incident on a lens parallel to the principal axis then it will converge to a single point F2 on principal axix as shown in the figure. This point is known as focus or second focal point. Similarly a diverging beam from point F1 when incidented on lense then it will travel along a parallel direction to principal axis and the point is know as first focal point
Principal points
Let us consider a lense having principal focus f1 and f2. Consider a ray of light AB parallel to principal axis be incident on lens. Then it will refreact along BC and finally emerge out along CF2 i.e principal focus II. Now Producing AB in initial direction and CF2 in backward direction. Then they will intersect in one point H2. A plane passing through H2 and perpendicular to principal axis is called principal plane and point of intersection betweeb this principal point and principal axis is called 2nd principal point
Similarly if we consider a ray of light F1D from first principal plane be incidented on lens and refract along DE and emerge along EG parallel to principal axis. Then point of intersection H1 producing F1D and EG in backward direction. A plane passing forn H1 and perpendicular to principal axis is called first principal plane. The Point of intersection of first plane and principal axis is called as first principal point
Nodal Point
NOdal point are pair of Conjugate points having unit angular magnification. This ,eans when an incident ray be incident on lense directed to either of nodal points and parallel to initial direction
By definition, Q1=Q2
So that P1N1=P2N2
H1H2=N1N2
If a parallel beam of light incident on a lens parallel to the principal axis then it will converge to a single point F2 on principal axix as shown in the figure. This point is known as focus or second focal point. Similarly a diverging beam from point F1 when incidented on lense then it will travel along a parallel direction to principal axis and the point is know as first focal point
Principal points
Let us consider a lense having principal focus f1 and f2. Consider a ray of light AB parallel to principal axis be incident on lens. Then it will refreact along BC and finally emerge out along CF2 i.e principal focus II. Now Producing AB in initial direction and CF2 in backward direction. Then they will intersect in one point H2. A plane passing through H2 and perpendicular to principal axis is called principal plane and point of intersection betweeb this principal point and principal axis is called 2nd principal point
Similarly if we consider a ray of light F1D from first principal plane be incidented on lens and refract along DE and emerge along EG parallel to principal axis. Then point of intersection H1 producing F1D and EG in backward direction. A plane passing forn H1 and perpendicular to principal axis is called first principal plane. The Point of intersection of first plane and principal axis is called as first principal point
Nodal Point
NOdal point are pair of Conjugate points having unit angular magnification. This ,eans when an incident ray be incident on lense directed to either of nodal points and parallel to initial direction
By definition, Q1=Q2
So that P1N1=P2N2
H1H2=N1N2
Codition for achromatism
Combination of two or more lenses such that there will not be a chromatic aberration is known as achromatic lense and the condition is called as achromatism
Mathmatically, df = fr-fv = 0
Now let us take two lenses having focal length f1&f2 and are separated by a distance x. Then theequivalent focal length of combination will be
1/f=1/f1 + 1/f2 - x/(f1*f2)
Diffrentiating this equation we have
-df/f2 + -df1/f1^2 - df2/f2^2 - x(-df1/f1^2f2 - df2/f1f2^2)
But we have
-df1/f1=w1 and -df2/f2 = w2
where w1 and w2 are the dispersive power of 1st and 2nd ;ems
-df/f2 = w1/f1 + w2/f2 - x( w1/f1f2 + w2/f1/f2)
For achromatism df = 0
w1/f1 + w2/f2 = x(w1/f1f2 + w2/f1f2) = 0
When two lense are kept in contact then x = 0
w1/f1 + w2/f2 = 0
f1/f2 = -w1/w2
The negative sign here indicates combination of lense must be of one convex and another concave
Mathmatically, df = fr-fv = 0
Now let us take two lenses having focal length f1&f2 and are separated by a distance x. Then theequivalent focal length of combination will be
1/f=1/f1 + 1/f2 - x/(f1*f2)
Diffrentiating this equation we have
-df/f2 + -df1/f1^2 - df2/f2^2 - x(-df1/f1^2f2 - df2/f1f2^2)
But we have
-df1/f1=w1 and -df2/f2 = w2
where w1 and w2 are the dispersive power of 1st and 2nd ;ems
-df/f2 = w1/f1 + w2/f2 - x( w1/f1f2 + w2/f1/f2)
For achromatism df = 0
w1/f1 + w2/f2 = x(w1/f1f2 + w2/f1f2) = 0
When two lense are kept in contact then x = 0
w1/f1 + w2/f2 = 0
f1/f2 = -w1/w2
The negative sign here indicates combination of lense must be of one convex and another concave
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