Question

A glass plate of thickness $t$ is placed in front of one of the slits in a Young's double slit experiment. It is found that the interference pattern is shifted through a distance $d$. If this whole setup is immersed in water, the shift in pattern will be:
$[\text{Given: Refractive index of glass,}{\mu }_g=\frac{3}{2},\text{Refractive index of water}{\mu }_w=\frac{4}{3}]$

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

Answer 1

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

The shift in interference pattern is given by, $S=\frac{D}{d}(\mu -1)t$

$\therefore \frac{S_1}{S_2}=\frac{{\mu }_1-1}{{\mu }_2-1}$

Here, ${\mu }_1={\mu }_g=\frac{3}{2}=1.5$, and, ${\mu }_2=\frac{{\mu }_g}{{\mu }_w}=\frac{9}{8}=1.125$

$\therefore \frac{S_1}{S_2}=\frac{1.5-1}{1.125-1}$

$\frac{d}{S_2}=\frac{0.5}{0.125}$

$\therefore S_2=0.25d=\frac{d}{4}$

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