In the arrangement shown in Fig., slits `S_(1)` and `S_(4)`are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of `S_(1) S_(2)` and `S_(3) S_(4)`.

If a hole is made at O' on AO' O and the slit `S_(4)` is closed, then the ratio of the maximum to minimum observed on screen at O , if `O' S_(3)` is equal to `(lambda D)/(4d)`, is

Options

A. 1
B. infinity
C. 34
D. 4

Answer

c

Explanation

Intensity at `S_(3) = A_(0)^(2) + A_(0)^(2) + 2A_(0)^(2) cos phi`
where `phi = (2pi)/(beta) (lambda D)/(4d) = (2 pi d)/(lambda D) xx (lambdaD)/(4d) = (pi)/(2)`
`I_(s_(3)) = A_(0)^(2) (2 + 2 cos ((pi)/(2)))`
`= 2 A_(0)^(2) xx cos^(2) ((pi)/(4)) = (4)/(2) A_(0)^(2) = 2 A_(0)^(2)`
Amplitude of wave at `S_(3), A_(3) = sqrt 2 A_(0)`
Maximum intensity at O' `= (A_(0) + sqrt(2 A_(0)))^(2)`
Maximum intensity at O `= (sqrt(2 A_(0)) - A_(0))^(2)`
`(I_(max))/(I_(min)) = ((sqrt 2 + 1)/(sqrt 2 - 1))^(2) = (sqrt 2 + 1)^(4) = 34`