Question

In Young's double-slit experiment, the point source $S$ is placed slightly off the central axis, as shown in figure. If $\lambda =500\text{nm}$, then match the following

Column $\left(\text{I}\right)$	Column $\left(\text{II}\right)$
$\left(A\right)$ Nature and order of interference at point $P$, $OP=10\text{mm}$	$1.$ Bright fringe of order $80$
$\left(B\right)$ Nature and order of interference at point $O$	$2.$ Bright fringe of order $262$
$\left(C\right)$ If a transparent paper (refractive index $\mu =1.45$) of thickness $t=0.02\text{mm}$ is pasted on $S_1$, i.e., one of the slits, the nature and order of interference at point $P$	$3.$ Bright fringe of order $62$
$\left(D\right)$ After inserting the transparent paper in front of Slit $S_1$, the nature and order of interference at $O$.	$4.$ Bright fringe of order $280$

• A. $A\to 1;B\to 3;C\to 4;D\to 2$
• B. $A\to 1;B\to 2;C\to 3;D\to 4$
• C. $A\to 3;B\to 1;C\to 2;D\to 4$
• D. $A\to 4;B\to 1;C\to 2;D\to 3$

Answer 1

The correct option is D $A\to 4;B\to 1;C\to 2;D\to 3$


The optical path difference between the two waves arriving at $P$ is

$\delta =(S{S}_2+S_{2}P)-(S{S}_1+S_{1}P)$

$=(S{S}_2-S{S}_1)+(S_{2}P-S_{1}P)$

$=d\sin {\theta }_0+d\sin \theta =d\tan {\theta }_0+d\tan \theta$

$\because {\theta }_0\&\theta$ are small

$\delta =\frac{d{y}_0}{D_1}+\frac{dy}{D_2}$

$d=20\text{mm},y_0=2\text{mm},D_2=2\text{m},D_1=1\text{m},y=10\text{mm}$

$\therefore \delta =\frac{20\times 2}{1000}+\frac{20\times 10}{2000}=0.14\text{mm}$

For a bright fringe, $\delta =n\lambda$

$\Rightarrow n=\frac{\delta }{\lambda }=\frac{0.14}{0.5\times {10}^{-3}}=280$

Hence, $\left(\text{A}\right)$ matches with $\left(4\right)$.

At the origin $O,{\delta }^{\prime }=\frac{d{y}_0}{D_1}=0.04\text{mm}$

$\therefore n^{\prime }=\frac{{\delta }^{\prime }}{\lambda }=\frac{0.04}{0.5\times {10}^{-3}}=80$

Hence, $\left(B\right)$ matches with $\left(1\right)$

Due to transparent paper, the change in the optical path is

$(\mu -1)t=(1.45-1)\left(0.02\right)\text{mm}=0.009\text{mm}$

$S^{\prime }=0.14-0.009=0.131\text{mm}$

$\Rightarrow n=\frac{0.131}{0.5\times {10}^{-3}}=262$

Hence, $\left(C\right)$ matches with $\left(2\right)$

Due to transparent paper, the path difference at $O$

$\delta ={\delta }^{\prime }-(\mu -1)t=(0.04-0.009)\text{mm}=0.031\text{mm}$

$\Rightarrow n=\frac{0.031}{0.5\times {10}^{-3}}=62$

Hence, $\left(D\right)$ matches with $\left(3\right)$

​​​​​​​Therefore, option $\left(\text{D}\right)$ is the correct answer.