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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_

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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,$\phi/2 = 60^{\circ} = \pi/3$ radians,which means $\phi = 2\pi/3$.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 also given by $\Delta = d \sin 	heta$.Equating the two expressions for $\Delta$,we get $d \sin 	heta = \lambda/3$.Therefore,$\sin 	heta = \lambda/(3d)$,which gives $	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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