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)` is
`I_(S_(3)) = A_(0)^(2)+ A_(0)^(2) + 2A_(0)^(2) cos phi`
Where `phi = (2pi)/(beta)xx(lambda D)/(4d) = (2pi)/(lambda D)xxdxx(lambda D)/(4d) = (pi)/(2)`
`I_(S_(3)) = 2A_(0)^(@)+2A_(0)^(2)cos phi`
`= 2A_(0)^(2)[1+cos phi] = 2 A_(0)^(2) [2cos^(2) (phi//2)]`
`I_(S_(3))= 4 A_(0)^(2) cos^(2) (phi//2) = 4A_(0)^(2)cos^(2) (pi//4)`
I_(S_(3)) = 4A_(0)^(2)xx(1)/(2) = 2 A_(0)^(2)`
Amplitude of wave at `S_(3) : A_(S_(1)) = sqrt(2) A_(0)`
Maximum intensity at `O' = (A_(0)+ sqrt(2)A_(0))`
Minimum intensity at `O = (sqrt(2) A_(0) -A_(0)-A_(0))^(2)`
`(I_(max))/(I_(min)) = [(A_(0)(sqrt(2)+1))/(A_(0)(sqrt(2)-1))]^(2) = [((sqrt(2)+1))/((sqrt(2)-1))xx(sqrt(2)+1)/(sqrt(2)-1)]^(2)`
`(I_(max))/(I_(min)) = [((sqrt(2)+1)^(2))/((sqrt(2)-1))]^(2) = (sqrt(2) + 1)^(4)`
`= (2+1+2sqrt(2))^(2) = (3+2sqrt(2))^(2)`
`= (3+2xx1.414)^(2) = (3+2.828)^(2)`
` = (5.828)^(2) = 34 , (I_(max))/(I_(min)) = 34`