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(C) The intensity at any point in a double-slit experiment is given by $I = I_{m} \cos^{2}(\phi / 2)$,where $I_{m}$ is the maximum intensity and $\phi

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

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(C) The intensity at any point in a double-slit experiment is given by $I = I_{m} \cos^{2}(\phi / 2)$,where $I_{m}$ is the maximum intensity and $\phi$ is the phase difference.Given that $I = I_{m} / 4$,we have:$I_{m} / 4 = I_{m} \cos^{2}(\phi / 2)$$\cos^{2}(\phi / 2) = 1/4$$\cos(\phi / 2) = 1/2 = \cos(\pi / 3)$$\phi / 2 = \pi / 3 \Rightarrow \phi = 2\pi / 3$.The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = (2\pi / \lambda) \Delta x$.Since $\Delta x = d \sin 	heta$,we have $\phi = (2\pi / \lambda) d \sin 	heta$.Equating the two expressions for $\phi$:$(2\pi / \lambda) d \sin 	heta = 2\pi / 3$$\sin 	heta = \lambda / (3d)$$	heta = \sin^{-1}(\lambda / 3d)$.

In Young's double-slit experiment,the intensity at a point is $1/4^{th}$ of the maximum intensity. The angular position of this point is

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