Consider the situation shown in figure. The two slite `S_1 and S_2`
placed symmetrically around the centre line are illuminated by a monochromatic
light of wavelength lambda. The separation between the slits is d. The light transmitted
by the slits falls on a screen `M_1` placed at a distance D from the slits. The slit `S_3` is
at the centre line and the slit `S_4` is at a distance y form `S_3`. Another screen `M_2` is
placed at a further distance D away from `M_1`. Find the ration of the maximum to
minimum intensity observed on `M_2` if y is equal to `(dltltD)`.
`
`(a) (lambdaD)/(2d) (b)(lambdaD)/d (c) (lambdaD)/(4d)`

Answer

A, B, C, D

Explanation

`S_(1)S_(3) = S_(2)S_(3)`
`:. Deltax "at" S_3` is zero. Therefore, intensity at `S_3` will be `4I_0`. Let us call it `I_1`. Thus,
`I_(S_3) = I_1 = 4I_0`
Now, `I_(S_4)` (or `I_2)` depends on the value of y.
(a) When y= `(lambdaD)/(2d)`
` Deltax = (yd)/D = lambda/2`
`:. I_(S_4) = I_2 = 0 `
or ` I_(max)/I_(min) = ((sqrtI_1+sqrtI_(2))/(sqrtI_(1) - sqrtI_(2)))^2 = 1`
`(b) When y= (lambdaD)/d `
`Deltax = (yd)/D = lambda`
` :. I_(S_4) = I_2 = 4I_0`
or ` I_(max)/I_(min) = ((sqrtI_(1) + sqrtI_(2))/(sqrtI_(1) - sqrt_(2)))^2 = oo`
( c) When `y=(lambdaD)/(4d)`
`rArr Deltaphi=(yd)/D=lambda/4`
`:. I_(S_4) = I_2 = 4I_0 (cos^2) phi/2 = 2I_0`
or ` I_(max)/I_(min) = (((sqrtI_(1)+sqrtI_(2))/(sqrtI_(1) - sqrtI_(2)))^2 = 34` .