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(C) The intensity at any point in an interference pattern is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.Given $I = (1/

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$\sin^{-1}(\lambda /3d)$

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(C) The intensity at any point in an interference pattern is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.Given $I = (1/4) I_{max}$,we have $(1/4) I_{max} = I_{max} \cos^2(\phi/2)$.This simplifies to $\cos^2(\phi/2) = 1/4$,so $\cos(\phi/2) = 1/2$.Thus,$\phi/2 = 60^{\circ} = \pi/3$ radians,which means $\phi = 120^{\circ} = 2\pi/3$ radians.The path difference $\Delta$ is related to the phase difference by $\Delta = (\lambda / 2\pi) 	imes \phi = (\lambda / 2\pi) 	imes (2\pi/3) = \lambda/3$.For small angles,the path difference is $\Delta = d \sin 	heta$,so $\sin 	heta = \Delta / d = \lambda / (3d)$.Therefore,the angular position is $	heta = \sin^{-1}(\lambda / 3d)$.

In Young's double-slit experiment,the intensity at a point is $(1/4)$ of the maximum intensity. The angular position of this point is:

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