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

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
  1. $\sin ^{-1}\left(\frac{\lambda}{D}\right)$
  2. $\sin ^{-1}\left(\frac{\lambda}{2 d}\right)$
  3. $\sin ^{-1}\left(\frac{\lambda}{3 d}\right)$
  4. $\sin ^{-1}\left(\frac{\lambda}{4 d}\right)$

Solution

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

Asked in: MHT CET 2024 (16 May Shift 1)

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