In Young's double-slit experiment, the intensity of light in front of one of the slits on a screen is `I_(0)//2` where `I_(0)` is the maximum intensity. The distance between the slits is `5lambda` where `lambda` is the wavelength of monochromatic light. How far away is the screen from the slit?

Options

A. `20lambda`
B. `25lambda`
C. `40lambda`
D. `50lambda`

Answer

D

Explanation

`(dx)/(D) = Deltax`
`2I = I +I +2I cos theta`
`cos theta = 0`
`phi = pi//2`
`(2pi)/(lambda) Delta x = (pi)/(2)`
`Delta x = (lambda)/(4)`
`(d xx x)/(D) = (lambda)/(4)`
`(5lambda)/(D) xx (5lambda)/(2) = (lambda)/(4)`
`D = 50 lambda`