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(C) The intensity at any point on the screen is given by $I = I_0 \cos^2 \left( \frac{\delta}{2} \right)$,where $I_0$ is the maximum intensity and $\d

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

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(C) The intensity at any point on the screen is given by $I = I_0 \cos^2 \left( \frac{\delta}{2} ight)$,where $I_0$ is the maximum intensity and $\delta$ is the phase difference.Given $I = \frac{I_0}{4}$,we have $\frac{I_0}{4} = I_0 \cos^2 \left( \frac{\delta}{2} ight)$.$\cos^2 \left( \frac{\delta}{2} ight) = \frac{1}{4} \implies \cos \left( \frac{\delta}{2} ight) = \frac{1}{2}$.$\frac{\delta}{2} = \frac{\pi}{3} \implies \delta = \frac{2\pi}{3}$.The path difference $\Delta x$ is related to phase difference by $\delta = \frac{2\pi}{\lambda} \Delta x$.$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x \implies \Delta x = \frac{\lambda}{3}$.For angular position $	heta$,the path difference is $\Delta x = d \sin 	heta$.$\frac{\lambda}{3} = d \sin 	heta \implies \sin 	heta = \frac{\lambda}{3d}$.Therefore,$	heta = \sin^{-1} \left( \frac{\lambda}{3d} ight)$.

In Young's double-slit experiment,the intensity at a point on the screen is $I_0/4$. Find the angular position of this point.

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