In young's double-slit experiment set up, sources S of wavelength 50 nm illumiantes two slits `S_(1)` and `S_(2)` which act as two coherent sources. The sources S oscillates about its own position according to the equation `y = 0.5 sin pi t`, where y is in nm and t in seconds. The minimum value of time t for which the intensity at point P on the screen exaclty in front of the upper slit becomes minimum is

Options

A. 1s
B. 2s
C. `0.5s`
D. `1.5s`

Answer

C

Explanation

`y' = (d)/(2)` at point P exactly opposite to `S_(1)`
`Delta x = (ys)/(D) + (d^(2))/(2D')`
For minimum intensity
`Delta x = (2n - 1) (lambda)/(2)`
`0.5sin pi t xx10^(-6) + (10^(-6))/(2xx2) = (2n-1) (1)/(2)`
At `t = 0`, at point 'P' however minimum is formed. again to get minimum after `t = 0, n = 2`
`0.5sin pi txx10^(-6)+0.25xx10^(-6)= (3)/(2)xx5xx10^(-7)`
`(1)/(2)sin pi t = 0.75 - 0.25`
`sin pi t = 1`
`t = (1)/(2) implies t = 0.5s`