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)`.
` (## DCP_V05_C32_S01_022_Q01.png" width="80%">
` (a) `(lambdaD)/(2d)` (b)`(lambdaD)/d` (c) `(lambdaD)/(4d)`

Options

A. 1
B. `1//2`
C. `3//2`
D. 2

Answer

a

Explanation

`z = (lambda D)/(2 d)`
At `S_(4): (Delta x)/(4) = (z)/(D)`
`implies Delta x = (lambda D)/(2 d) = (lambda)/(2)`
Hence, minima at `S_(4)` and maxima at `S_(3)` (intensity `4 I_(0)`).
Hence,
`(I_(max))/(I_(min)) = 1`