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

• A. $co{s}^{-1}\left(\frac{1}{12}\right)$
• B. $si{n}^{-1}\left(\frac{1}{12}\right)$
• C. $ta{n}^{-1}\left(\frac{1}{12}\right)$
• D. $si{n}^{-1}\left(\frac{3}{5}\right)$

Answer 1

The correct option is A

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

$I=I_{0}co{s}^2\left(\frac{\varphi }{2}\right)$

But $\frac{I}{I_0}\left(given\right)orco{s}^2\left(\frac{\varphi }{2}\right)=\frac{3}{4}$

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

$(\because d=2\lambda )$

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

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