Question
In Young’s double slit experiment intensity at a point is $$\left( {\frac{1}{4}} \right)$$ of the maximum intensity. Angular position of this point is
A.
$${\sin ^{ - 1}}\left( {\frac{\lambda }{{d}}} \right)$$
B.
$${\sin ^{ - 1}}\left( {\frac{\lambda }{{2\,d}}} \right)$$
C.
$${\sin ^{ - 1}}\left( {\frac{\lambda }{{3\,d}}} \right)$$
D.
$${\sin ^{ - 1}}\left( {\frac{\lambda }{{4\,d}}} \right)$$
Answer :
$${\sin ^{ - 1}}\left( {\frac{\lambda }{{3\,d}}} \right)$$
Solution :
$$\eqalign{ & I = {I_{\max }}{\cos ^2}\frac{{\pi d\sin \theta }}{\lambda } \cr & \Rightarrow \,\,\frac{{{I_{\max }}}}{4} = {I_{\max }}{\cos ^2}\left( {\frac{{\pi d\sin \theta }}{\lambda }} \right) \cr & \therefore \,\,\cos \frac{{\pi d\sin \theta }}{\lambda } = \frac{1}{2} \cr & \therefore \,\,\frac{{\pi d\sin \theta }}{\lambda } = \frac{\pi }{3} \cr & \therefore \,\,\theta = {\sin ^{ - 1}}\left( {\frac{\lambda }{{3\,d}}} \right) \cr} $$
$$\eqalign{ & I = {I_{\max }}{\cos ^2}\frac{{\pi d\sin \theta }}{\lambda } \cr & \Rightarrow \,\,\frac{{{I_{\max }}}}{4} = {I_{\max }}{\cos ^2}\left( {\frac{{\pi d\sin \theta }}{\lambda }} \right) \cr & \therefore \,\,\cos \frac{{\pi d\sin \theta }}{\lambda } = \frac{1}{2} \cr & \therefore \,\,\frac{{\pi d\sin \theta }}{\lambda } = \frac{\pi }{3} \cr & \therefore \,\,\theta = {\sin ^{ - 1}}\left( {\frac{\lambda }{{3\,d}}} \right) \cr} $$