Question

In the arrangement shown in the figure, slits $S_3$ and $S_4$ are having a variable separation $Z$. Point $O$ on the screen is at the common perpendicular bisector of $S_1{S}_2$ and $S_3{S}_4$.
When $Z=\frac{\lambda D}{2d}$ the intensity measured at $O$ is $I_0$. The intensity at $O$ when $Z=\frac{2\lambda D}{d}$ is $n{I}_0$. Find the value of $n$.

• A. 2.00
• B. 2
• C. 2.0

Answer 1

Answer: (A), (B), (C)

Intensity at $O$ is proportional to intensity at $S_3$ and $S_4$.

Let the intensity at $S_1\text{and}{S}_2$ be $I_1$ and the intensity at $S_3\text{and}{S}_4$ be $I_2$.

Let the intensity at $O$ be $I_3$.

As, $O$ is at the perpendicular bisector,
so the path difference between $S_3\text{and}{S}_4$ at $O$ is zero.

Hence, $I_3=4I_2.......\left(1\right)$

By resultant intensity formula,
$I_2=4I_1{\cos }^2\left(\frac{\varphi }{2}\right)$

The phase difference $\varphi$ at $S_3\text{and}{S}_4$ is given by

$\varphi =\left(\frac{2\pi }{\lambda }\right)\left(\Delta x\right)=\left(\frac{2\pi }{\lambda }\right)\left(\frac{yd}{D}\right)$.

Here, $y=\frac{\text{z}}{2}$

$\Rightarrow \varphi =\left(\frac{2\pi }{\lambda }\right)\left(\frac{\text{z}d}{2D}\right)$.

At, $\text{z}=\frac{\lambda D}{2d}$

$\varphi =\frac{\pi }{2}\Rightarrow I_2=2I_1$

$\Rightarrow I_3=8I_1$

Given that,

At $\text{z}=\frac{\lambda D}{2d}\Rightarrow I_3=I_0$

$\therefore I_1=\frac{I_0}{8}$

Now, when $\text{z}=\frac{2\lambda D}{d}$

$\varphi =2\pi \Rightarrow I_2=4I_1\Rightarrow I_3=16I_1$

$\therefore I_3=2I_0$

Comapring this with $I_3=n{I}_0$, we get $n=2$