In a Young's double slit experiment, the light sources is at distance `l_1 = 20mu m ` and `l_2 = 40 mu m` form the main slit. The light of wavelength lambda = 500nm is incident on slits separated at a distance `10mu m `. A screen is placed at a distance D = 2 away from the slits as shown in figure. Find
(a) the values of `theta` relative to the central line where maxima appear on the screen?
(b) how many maxima will appear on the screen?
what should be the minimum thickness of a slab of refractive index 1.5 placed on the path of one of the ray so that minima occurs at C ?
` .

Explanation

(a) The optical path difference between the beams arriving at P,
` Deltax = (l_2 - l_1) + dsin theta`
The condition for maximum intensity is `[n = 0, +-1, +-2 ………….]`
`Deltax = nlambda`
Thus, ` sintheta = 1/d [Deltax - (l_2 - l_1)] = 1/d [nlambda -(l_2 -l_1)]`
`= 1/(10 xx (10^-6))[n xx 500 xx 10^(-9) xx 20 xx 10^(-6)] `
` = 2 [ (n/40) -1] `
Hence, `theta = sin^-1 [2((n/40)-1)]`
` (b) |sin theta| le1 `
`:. (-1le 2[(n/40)-1]le1)`
or ` -20 le(n-40)le20`
or ` 20le n le60`
Hence, number of maximas = 60-20 = 40
(c) At C, phase difference ` phi = (2pi)/lambda(l_2 - l_1) = ((2pi) / (500 xx 10^-9)) (20 xx 10^-6)`
` = 80 pi `
Hence, maximum intensity will appear at C. For minimum intensity at C,
`(mu-1)t = lambda/2 `
or l ` t= lambda/(2(mu-1)) = (500 xx 10^-9)/(2 xx 0.5) = 500nm ` .