In Young's double slit experiment the intensity of light on the screen where the 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. `I//4`
B. `I//2`
C. `I`
D. zero

Answer

B

Explanation

`I=4I^(') cos^(2) (phi//2)`
`I_(R)=4I^(') cos^(2)(phi//2)`
`I_(R)/I=(cos^(2)(phi^(')//2))/(cos^(2) (phi//2))`
`phi=(2pi)/lambdaDeltax=(2pi)/lambda.lambda=2pi`
`phi^(')=(2pi)/lambda.lambda/4=pi/2`
`I_(R)/I=(cos^(2)(pi//4))/(cos^(2)(pi))=(1//2)/1=1/2`
`I_(R)=I//2`