At two point P and Q on screen in Young's double slit experiment, waves from slits `S_(1)` and `S_(2)` have a path difference of 0 and `(lamda)/(4)` respectively. The ratio of intensities at P and Q will be:

Options

A. `3 : 2`
B. `2 : 1`
C. `sqrt(2) : 1`
D. `4 : 1`

Answer

B

Explanation

At pint `P` on the screen, path diff. between two waves reaching there, `x = 0`.
Therefore, phase diff. `phi = (2 pi)/(lambda) Delta x = 0`.
Here, `I_(1) = I_(2) = I_(0)` (say)
As `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi`
`:. I_(P) = I_(0) + I_(0) + 2 I_(0) cos 0 = 4 I_(0)`.
At point `Q` on the screen, path diff. between two waves reaching there `x = lambda//4`. Therefore phase diff.
`phi = (2 pi)/(lambda) xx (lambda)/(4) = (pi)/(2)`
`I_(Q) = I_(0) + I_(0) + 2 I_(0) "cos" (pi)/(2) = 2 I_(0)`
`:. (I_(p))/(I_(Q)) = (4 I_(0))/(2 I_(0)) = (2)/(1)`