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
$I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$
But $\frac{I}{I_0}=\frac{3}{4}$ (given)
or ${\cos }^2\left(\frac{\varphi }{2}\right)=\frac{3}{4}$
or $\cos \frac{\varphi }{2}=\frac{\sqrt{3}}{2}$
or $\varphi =60^{\circ }=\frac{\pi }{3}$
Phase difference, $\varphi =\frac{2\pi }{\lambda }\times$ path difference

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