The maximum intensity in young's double-slit experiment is `I_(0)`. Distance between the slit is `d = 5 lambda`, where `lambda` is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance `D = 10 d`?

Options

A. `I_(0)`
B. `(I_(0))/(4)`
C. `3/4I_(0)`
D. `(I_(0))/(2)`

Answer

D

Explanation

`Deltax=(yd)/D=(d/2xxd)/D`
`=(d^(2))/(2D)=((5lambda)^(2))/(2xx50 lambda)=(lambda)/4`
`phi=(2pi)/(lambda) Deltax=90^(@)`
`I=I_(max)cos^(2) phi//2=I_(0)//2`