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, Now IN triangle S2S2N
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