In the Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is `lambda` is `K` (`lambda` being the wave length of light used).The intensity at a point where the path difference is `lambda/4`, will be

Options

A. `K//2`
B. Zero
C. K
D. `k//4`

Answer

A

Explanation

Phase difference `(Deltaphi)= (2jpi)/(lambda) xx "Path difference"`
and Intensity (I)= `4I_(0)cos^(2)(phi)/(2)`
When path difference `(Deltax)= lambda`
`Deltaphi= (2pi)/(lambda) xx lambda= 2pi`
`I_("max")= 4I_(0)= K implies 2I_(0)= (K)/(2)`
When path difference `(Deltax)= (lambda)/(4)`
`Deltaphi= (2pi)/(lambda) xx (lambda)/(4) = (pi)/(2)`
`I= 4I_(0) cos^(2)(pi)/(4)= 2I_(0) :. I= (K)/(2)`