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(B) The maximum intensity is given by $I_{\max} = 4I_0 = I$,where $I_0$ is the intensity of each individual slit.The intensity at any point on the scr

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

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(B) The maximum intensity is given by $I_{\max} = 4I_0 = I$,where $I_0$ is the intensity of each individual slit.The intensity at any point on the screen is given by $I_p = 4I_0 \cos^2\left(\frac{\phi}{2}ight)$,where $\phi$ is the phase difference.Given that the intensity at the required point is $I_p = \frac{I}{4}$,we substitute $I = 4I_0$:$\frac{4I_0}{4} = 4I_0 \cos^2\left(\frac{\phi}{2}ight)$$I_0 = 4I_0 \cos^2\left(\frac{\phi}{2}ight)$$\cos^2\left(\frac{\phi}{2}ight) = \frac{1}{4}$$\cos\left(\frac{\phi}{2}ight) = \frac{1}{2}$This implies $\frac{\phi}{2} = \frac{\pi}{3}$,so $\phi = \frac{2\pi}{3}$.The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.For a double slit experiment,the path difference is $\Delta x = d \sin 	heta$.Therefore,$\frac{2\pi}{3} = \frac{2\pi}{\lambda} (d \sin 	heta)$.Solving for $\sin 	heta$ gives $\sin 	heta = \frac{\lambda}{3d}$.Thus,the angular position is $	heta = \sin^{-1}\left(\frac{\lambda}{3d}ight)$.

Column-$I$ (Molecules)	Column-$II$ (Number of lone pairs on central atom)
$A$. $NH_3$	$1$. Two
$B$. $H_2O$	$2$. Three
$C$. $XeF_2$	$3$. Zero
$D$. $CH_4$	$4$. Four
	$5$. One

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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