In Young's double slit experiment intensity at a point is `((1)/(4))` of the maximum intersity. Angular position of this point is

Options

A. `sin^(-1)(lambda//d)`
B. `sin^(-1)(lambda//2d)`
C. `sin^(-1)(lambda//3d)`
D. `sin^(-1)(lambda//4d)`

Answer

C

Explanation

`I=4I^(') cos^(2) (phi//2)=I_(m) cos^(2)(phi//2)`
`I_(m)/4=I_(m) cos^(2)(phi//2) implies 1/4=cos^(2)(phi//2)`
`cos(phi//2)=1/2=cos(pi//3)`
`phi=(2pi)/3`
`phi=(2pi)/lambdaDeltax=(2pi)/lambda(d sin theta)=(2pi)/3`
`sin theta=lambda/(3d) implies theta=sin^(-1)(lambda//3d)`