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

For net intensity `I=4I_(0) "cos"^(2) phi/2`
Case I `K=4I_(0)cos^(2)((2pi)/2)(because phi=(2pi)/lambdaxxlambda)`
`K=4I_(2)cos^(2)(pi)rArrK=4I_(0)....(i)`
Case II `K=4I_(0)cos^(2)((pi//2)/2)(because phi=(2pi)/lambdaxxlambda/4)`
`K=4I_(0)cos^(2)(pi//4)rArrK=2I_(0)....(ii)`
Compring Eqs. (i) and (ii) we get `K=K//2`