Question

A monochromatic light of $\lambda =6000\mathring{\text{A}}$ is incident on the slits. The interference pattern is observed on a screen and central maxima is formed at point $O$. A thin glass plate of thickness $0.4\mu \text{m}$ and refractive index $1.5$ is placed in front of one of the slits. If the intensity at point $O$ is $I_0$ initially, what will be the intensity at $O$ after placing the glass plate?

• A. $\frac{I_0}{4}$
• B. $\frac{I_0}{\sqrt{2}}$
• C. $\frac{I_0}{2}$
• D. $I_0$

Answer 1

The correct option is A $\frac{I_0}{4}$

Let $I$ be the intensity of each slit.

The intensity at central maxima is, $=I_0=4I\Rightarrow I=\frac{I_0}{4}$

The path difference at $O$, after placing the glass plate is,
$\Delta x=(\mu -1)t=(1.5-1)\times 0.4\times {10}^{-6}$
$\therefore \Delta x=0.2\times {10}^{-6}\text{m}$

The phase difference is, $\varphi =\frac{2\pi }{\lambda }\Delta x$
$\therefore \varphi =\frac{2\pi }{6\times {10}^{-7}}\times 0.2\times {10}^{-6}=\frac{2\pi }{3}\text{rad}$
The resultant intensity is,
$I_R=4I{\cos }^2\frac{\varphi }{2}=I=\frac{I_0}{4}$

Hence, $\left(C\right)$ is the correct answer.