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

Options

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

Answer

C

Explanation

Given `I_(R)=(4I)/(4)=I` So `I=I+I+2Icosphi`
`cosphi=-(1)/(2)impliesphi=(2pi)/(3)`
Corresponding path difference `Deltax=(lamda)/(3)`
So `dsintheta=(lamda)/(3)impliestheta=sin^(-1)((lamda)/(3d))`