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 wavelength of light used). The intensity at a point where the path difference is `lambda//4` will be

Options

A. `K`
B. `K//4`
C. `K//2`
D. zero

Answer

C

Explanation

When path diff. is `lambda, I = I_(max) = K`.
When path diff. is `lambda//4`
`I = I_(max) cos^(2)phi//2`
`= K cos^(2)[(2pi)/(lambda).(lambda)/(4) xx (1)/(2)]`
`= K (cos pi//4)^(2) = K((1)/sqrt(2))^(2) = (K)/(2)`