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

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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 $I = I_{max} \cos^2(\phi / 2)$,where $\phi$ is the phase difference.Given $I = I_{max} / 4$,we have $I_{max} / 4 = I_{max} \cos^2(\phi / 2)$.This simplifies to $\cos^2(\phi / 2) = 1/4$,so $\cos(\phi / 2) = 1/2$.Thus,$\phi / 2 = 60^{\circ}$ or $\pi / 3$ radians,which means the phase difference $\phi = 120^{\circ}$ or $2\pi / 3$ radians.The path difference $\Delta$ is related to the phase difference by $\Delta = (\lambda / 2\pi) 	imes \phi$.Substituting $\phi = 2\pi / 3$,we get $\Delta = (\lambda / 2\pi) 	imes (2\pi / 3) = \lambda / 3$.For small angles,the path difference $\Delta = d \sin 	heta$,where $d$ is the slit separation.Therefore,$d \sin 	heta = \lambda / 3$,which gives $\sin 	heta = \lambda / (3d)$.Hence,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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