MCQ
In Young's double slit experiment maximum intensity is $I$ then the angular position where the intensity becomes $\frac{I}{4}$ is
- A${\sin ^{ - 1}}\left( {\frac{\lambda }{d}} \right)$
- ✓${\sin ^{ - 1}}\left( {\frac{\lambda }{3d}} \right)$
- C${\sin ^{ - 1}}\left( {\frac{\lambda }{2d}} \right)$
- D${\sin ^{ - 1}}\left( {\frac{\lambda }{4d}} \right)$
