Question

In Young's double-slit experiment, the slits are horizontal. The intensity at a Point P 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 $2\lambda$)

Answer 1

Here $\frac{I}{I_0}=\frac{3}{4}$ (given)
$\Rightarrow co{s}^2\left(\frac{\varphi }{2}\right)=\frac{3}{4}$

or $cos\frac{\varphi }{2}=\frac{\sqrt{3}}{2}or\varphi ={60}^0=\frac{\pi }{3}$

Phase difference, $\varphi =\frac{2\pi }{\lambda }\times$ path difference is
$dcos\theta =2\lambda cos\theta$

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

$\therefore cos\theta =\frac{1}{12}$

$\theta =co{s}^{-1}\left[\frac{1}{12}\right]$