In Young's double slit experiment,the sli | vedclass

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(B) The intensity at any point in an interference pattern is given by $I = I_0 \cos^2\left(\frac{\phi}{2}\right)$,where $\phi$ is the phase difference

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$\sin^{-1}\left(\frac{1}{6}\right)$

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(B) The intensity at any point in an interference pattern is given by $I = I_0 \cos^2\left(\frac{\phi}{2}ight)$,where $\phi$ is the phase difference.
Given $I = \frac{I_0}{4}$,we have $\frac{I_0}{4} = I_0 \cos^2\left(\frac{\phi}{2}ight)$,which implies $\cos^2\left(\frac{\phi}{2}ight) = \frac{1}{4}$.
Taking the square root,$\cos\left(\frac{\phi}{2}ight) = \frac{1}{2}$,so $\frac{\phi}{2} = \frac{\pi}{3}$,which gives $\phi = \frac{2\pi}{3}$.
The path difference $\Delta x$ is related to the phase difference by $\Delta x = \frac{\lambda}{2\pi} \phi = \frac{\lambda}{2\pi} \left(\frac{2\pi}{3}ight) = \frac{\lambda}{3}$.
From the geometry of the setup,the path difference is $\Delta x = d \sin 	heta$. Given $d = 2\lambda$,we have $\frac{\lambda}{3} = 2\lambda \sin 	heta$.
Solving for $\sin 	heta$,we get $\sin 	heta = \frac{\lambda/3}{2\lambda} = \frac{1}{6}$.
Therefore,$	heta = \sin^{-1}\left(\frac{1}{6}ight)$.

In Young's double slit experiment,the slits are horizontal. The intensity at a point $P$ on the screen is $\frac{I_0}{4}$,where $I_0$ is the maximum intensity. If the distance between the two slits $S_1$ and $S_2$ is $d = 2 \lambda$,then the value of $\theta$ is:

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