Physics - Wave Optics Question with Solution | TestHub

The intensity at any point in Young's double slit experiment is given by:

$I=I_{\max }{\cos }^2\left(\frac{\varphi }{2}\right)$

where $\varphi$ is the phase difference between the interfering waves.

Given that $I=\frac{I_{\max }}{4}$, we have:

$\frac{I_{\max }}{4}=I_{\max }{\cos }^2\left(\frac{\varphi }{2}\right)$

${\cos }^2\left(\frac{\varphi }{2}\right)=\frac{1}{4}$

$\cos \left(\frac{\varphi }{2}\right)=\frac{1}{2}$

$\frac{\varphi }{2}=\frac{\pi }{3}$

$\varphi =\frac{2\pi }{3}$

The relation between phase difference $\varphi$ and path difference $\Delta x$ is:

$\varphi =\frac{2\pi }{\lambda }\Delta x$

$\frac{2\pi }{3}=\frac{2\pi }{\lambda }\Delta x$

$\Delta x=\frac{\lambda }{3}$

The path difference in terms of the angular position $\theta$ is given by $\Delta x=d\sin \theta$, where $d$ is the distance between the slits.

$d\sin \theta =\frac{\lambda }{3}$

$\sin \theta =\frac{\lambda }{3d}$

$\theta ={\sin }^{-1}\left(\frac{\lambda }{3d}\right)$

Answer: ${\sin }^{-1}\left(\frac{\lambda }{3d}\right)$