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(B) The resultant intensity in Young's double slit experiment is given by $I_R = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$,where $I = 4I_0$ is the maxim

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$\sin^{-1}\left(\frac{\lambda}{3d}\right)$

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(B) The resultant intensity in Young's double slit experiment is given by $I_R = 4I_0 \cos^2\left(\frac{\phi}{2}ight)$,where $I = 4I_0$ is the maximum intensity.Given the intensity at a point is $I' = \frac{I}{4} = \frac{4I_0}{4} = I_0$.Substituting this into the formula: $I_0 = 4I_0 \cos^2\left(\frac{\phi}{2}ight)$.$\cos^2\left(\frac{\phi}{2}ight) = \frac{1}{4} \implies \cos\left(\frac{\phi}{2}ight) = \frac{1}{2}$.This implies $\frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3}$.The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.Since $\Delta x = d \sin 	heta$,we have $\phi = \frac{2\pi}{\lambda} d \sin 	heta$.Equating the two expressions for $\phi$: $\frac{2\pi}{\lambda} d \sin 	heta = \frac{2\pi}{3}$.Solving for $\sin 	heta$: $\sin 	heta = \frac{\lambda}{3d}$.Therefore,$	heta = \sin^{-1}\left(\frac{\lambda}{3d}ight)$.

In Young's double slit experiment,if the maximum intensity is $I$,then the angular position where the intensity becomes $\frac{I}{4}$ is:

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