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

Options

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

Answer

C

Explanation

Path difference `=(dx)/(D)`
`=(dxx(d)/(2))/(10d)`
`=(d)/(20)=(5 lambda)/(20)`
`"sin" theta=(lambda)/(4)`
`:. sin theta=(2pi)/(4) [ :. lambda= 2pi]`
`:. sin theta=(pi)/(2)`
`:. theta=90^(@)`
From Malus, law,
`I=I_(0)"cos"^(2)(phi)/(2)=I_(0)cos^(2)45^(@)`
`:.I=I_(0)xx((1)/(sqrt(2)))^(2) :.I=(I_(0))/(2)`