In Young's double-slit experiment, the intensity at a point is...

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

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

\dfrac{I_0}{4} =$I_0 \cos^2 \left(\dfrac{\phi}{2}\right)$

$\cos\left(\dfrac{\phi}{2}\right) = \dfrac{1}{2}$

$\dfrac{\phi}{2} = \dfrac{\pi}{3} \Rightarrow \phi = \dfrac{2\pi}{3}$

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

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

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

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

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

$\theta = \sin^{-1} \left(\dfrac{\lambda}{3d}\right)<br><br>Answer:$\sin^{-1} \left(\dfrac{\lambda}{3d}\right)