Question

In Young’s double slit experiment , the slit are horizontal. The intensity at a point P as shown in figure is $\frac{3}{4}$where $I_{0}$is the maximum intensity. 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. $\sin^{- 1}\left( \frac{1}{12} \right)$

Answer 1

Answer: (A)

$\cos^{- 1}\left( \frac{1}{12} \right)$

Answer 2

: In young’s double slit experiment, intensity at as point is given by

$I_{} = I_{}\cos^{2}\left( \frac{\varphi}{2} \right)$

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

$d\cos\theta = 2\lambda\cos\theta$ $(\because d = 2\lambda)$

$\therefore\frac{\pi}{3} = \frac{2\pi}{\lambda}2\lambda\cos\theta\therefore\cos\theta = \frac{1}{12}$

$\therefore\theta = \cos^{- 1}{}\left( \frac{1}{12} \right)$