Question

A monochromatic light of $\lambda =500A^{\circ }$ is incident on two identical slits separated by a distance of $5\times {10}^{-4}\text{m}$. The interference pattern is seen on a screen placed at a distance of $1\text{m}$ from the plane of slits. A thin glass plate of thickness $1.5\times {10}^{-6}\text{m}$ and refractive index $\mu =1.5$ is placed between one of the slits and the screen. Find the intensity at the center of the slit now.

Answer 1

Take a point of distance $y$ from the center of the slit. Path difference between waves reaching this point.

$\Delta x=\frac{dy}{D}-(\mu -1)t$

For center of slit , $y-0$

So, $\Delta x=-(\mu -1)t$

Phase difference $\varphi =\frac{2\pi }{\lambda }\Delta x=-\frac{2\pi }{\lambda }(\mu -1)t$

Therefore, Intensity $I=4I_0{\cos }^2\left(\frac{\varphi }{2}\right)=4I_0{\cos }^2\left(\frac{\pi }{\lambda }\right(\mu -1\left)t\right)$

Substituting the given data gives,

$I=4I_0{\cos }^2\left[\frac{\pi (1.5-1)1.5\times {10}^{-6}}{500\times {10}^{-10}}\right]=4I_0{\cos }^2\left(\frac{3\pi }{2}\right)=0$