Question 278: Waves Optics

Question

In Young's double slit experiment, intensity at a point is $\left(\frac{1}{4}\right)$ of the maximum intensity. The angular position of this point is

Accepted Answer

For any point in interference pattern,
$$\begin{array}{ll} 
& \mathrm{I}=\mathrm{I}_{\max } \cos ^2 \frac{\phi}{2} \
	herefore \quad & \frac{\mathrm{I}_{\max }}{4}=\mathrm{I}_{\max } \cos ^2 \frac{\phi}{2} \
	herefore \quad & \cos ^2 \frac{\phi}{2}=\frac{1}{4} \
	herefore \quad & \cos \frac{\phi}{2}=\frac{1}{2} \
	herefore \quad & \frac{\phi}{2}=60^{\circ}=\frac{\pi}{3} \
	herefore \quad & \phi=\frac{2 \pi}{3}
\end{array}$$
We know that,
$\phi=\left(\frac{2 \pi}{\lambda}ight) \Delta x$, where $\Delta x$ is path difference.
$$\begin{array}{ll} 
& 	ext { and } \Delta x=d \sin 	heta \
	herefore & \frac{2 \pi}{3}=\frac{2 \pi}{\lambda}(\mathrm{~d} \sin 	heta) \
	herefore & \frac{\lambda}{3 \mathrm{~d}}=\sin 	heta \
	herefore & 	heta=\sin ^{-1}\left(\frac{\lambda}{3 \mathrm{~d}}ight)
\end{array}$$

Answer

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

Answer

$\sin ^{-1}\left(\frac{\lambda}{2 d}\right)$

Answer

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

Answer

$\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$

Waves Optics

In Young's double slit experiment, intensity at a point is \left(\frac{1}{4}\right) of the maximum intensity. The angular position of this point is

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