Question

In Young's double slit experiment, the slits are horizontal. The intensity at a point $P$ as shown in figure is $\frac{3}{4} I_{0}$, where $I_{0}$ is the maximum intensity. Then the value of $\theta$ is, (Given the distance between the two slits $S_{1}$ and $S_{2}$ is $2 \lambda$ )

• A. $\cos ^{-1}\left(\frac{1}{12}\right)$
• B. $\sin ^{-1}\left(\frac{1}{12}\right)$
• C. $\tan ^{-1}\left(\frac{1}{12}\right)$
• D. $\sin ^{-1}\left(\frac{3}{5}\right)$

Answer 1

Answer: (A)

In Young's double slit experiment, intensity at a point is given by
$I=I_{0} \cos ^{2}\left(\frac{\phi}{2}\right)$
But $\frac{I}{I_{0}}=\frac{3}{4}$ (given)
or $\cos ^{2}\left(\frac{\phi}{2}\right)=\frac{3}{4}$
or $\cos \frac{\phi}{2}=\frac{\sqrt{3}}{2}$
or $\phi=60^{\circ}=\frac{\pi}{3}$
Phase difference, $\phi=\frac{2 \pi}{\lambda} \times$ path difference

From the figure, path difference is
$d \cos \theta=2 \lambda \cos \theta \quad(\because d=2 \lambda)$
$\therefore \frac{\pi}{3}=\frac{2 \pi}{\lambda} 2 \lambda \cos \theta$
$\therefore \cos \theta=\frac{1}{12}$
$\theta=\cos ^{-1}\left(\frac{1}{12}\right)$