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(C) The intensity $I$ 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 =

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

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(C) The intensity $I$ at any point in an interference pattern is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.Given $I = I_{max}/4$,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,$\phi/2 = \pi/3$,which means the phase difference $\phi = 2\pi/3$.The relation between phase difference $\phi$ and path difference $\Delta x$ is $\phi = (2\pi/\lambda) \Delta x$.Substituting $\phi = 2\pi/3$,we get $2\pi/3 = (2\pi/\lambda) \Delta x$,which gives $\Delta x = \lambda/3$.For a point at angular position $	heta$,the path difference is $\Delta x = d \sin 	heta$.Therefore,$d \sin 	heta = \lambda/3$,which implies $\sin 	heta = \lambda/(3d)$.Hence,$	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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