In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of the point is:
A. \[{\sin ^{ - 1}}\left( {\dfrac{{2\lambda }}{d}} \right)\]
B. \[{\sin ^{ - 1}}\left( {\dfrac{\lambda }{{2d}}} \right)\]
C. \[{\sin ^{ - 1}}\left( {\dfrac{\lambda }{{3d}}} \right)\]
D. \[{\sin ^{ - 1}}\left( {\dfrac{\lambda }{d}} \right)\]

Hint: Use the formula for intensity in Young’s double slit experiment. Substitute the given value of intensity in the formula to get the phase difference. Use the relation between path difference and phase difference to determine the path difference. Use, \[\Delta l = d\sin \theta \] to equate the path difference you obtained.

Formula used:
The intensity at a point P when the phase difference between the two waves is \[\phi \] given as,
 \[I = {I_{\max }}{\cos ^2}\left( {\dfrac{\phi }{2}} \right)\]
Here, \[{I_{\max }}\] is the maximum intensity.
The relation between path difference and phase difference,
\[\Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta l\]
Here, \[\Delta l\] is the path difference

Complete step by step solution:
We know that in Young’s double slit experiment, the intensity at a point P when the phase difference between the two waves is \[\phi \] given as,
 \[I = {I_{\max }}{\cos ^2}\left( {\dfrac{\phi }{2}} \right)\]

Here, \[{I_{\max }}\] is the maximum intensity.

We have given that the intensity of the wave is \[I = \dfrac{{{I_{\max }}}}{4}\]. Therefore, substituting \[I = \dfrac{{{I_{\max }}}}{4}\] in the above equation, we get,
\[\dfrac{{{I_{\max }}}}{4} = {I_{\max }}{\cos ^2}\left( {\dfrac{\phi }{2}} \right)\]
\[ \Rightarrow \dfrac{1}{4} = {\cos ^2}\left( {\dfrac{\phi }{2}} \right)\]
\[ \Rightarrow \cos \left( {\dfrac{\phi }{2}} \right) = \dfrac{1}{2}\]
\[ \Rightarrow \phi = 2{\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]

We know that, \[{\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3}\]. Therefore, the above equation becomes,
\[\phi = \dfrac{{2\pi }}{3}\] …… (1)

We know the relation between path difference and phase difference,
\[\Delta \phi = \dfrac{{2\pi }}{\lambda }\Delta l\]
\[ \Rightarrow \Delta l = \dfrac{\lambda }{{2\pi }}\Delta \phi \] …… (2)

Here, \[\Delta \phi \] is the phase difference and \[\Delta l\] is the path difference.

We also have the relation from Young’s double slit experiment,
\[\Delta l = d\sin \theta \] …… (3)

Here, d is the distance between the slits and \[\theta \] is the angular position.

Equating equation (2) and (3), we get,
\[\dfrac{\lambda }{{2\pi }}\Delta \phi = d\sin \theta \]

Substituting \[\Delta \phi = \dfrac{{2\pi }}{3}\] from equation (1) in the above equation, we get,
\[\left( {\dfrac{\lambda }{{2\pi }}} \right)\left( {\dfrac{{2\pi }}{3}} \right) = d\sin \theta \]
\[ \Rightarrow \dfrac{\lambda }{3} = d\sin \theta \]
\[ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{\lambda }{{3d}}} \right)\]

So, the correct answer is “Option C”.

Note:
This solution solely depends on the formulae we derived in Young’s double slit experiment. Therefore, students should remember all the formulae in Young’s double slit experiment. Note that \[\theta \] is the angle made by the point P with the line joining the centre of two slits and the position of the first bright fringe.