Physics - Wave Optics Question with Solution | TestHub

The intensity at a point in Young's double-slit experiment is given by $I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$, where $I_0$ is the maximum intensity and $\varphi$ is the phase difference.

Given $I=\frac{I_0}{4}$

$\frac{I_0}{4}=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

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

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

The phase difference $\varphi$ is related to the path difference $\Delta x$ by $\varphi =\frac{2\pi }{\lambda }\Delta x$

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

The path difference is also given by $\Delta x=d\sin \theta$

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

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

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

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