Question

In a Young's double slit experiment, the distance between the slits & the screen is $100\text{cm}.$ For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width $0.25\text{mm}$. When the distance between the slits is increased by $1.2\text{mm},$ the fringe width decreased to $2/3$ of the original value. In the final position, a thin glass plate of refractive index $1.5$ is kept in front of one of the slits & the shift of central maximum is observed to be $20$ times the initial fringe width. Find the thickness of the plate.

• A. $24\mu \text{m}$
• B. $20\mu \text{m}$
• C. $30\mu \text{m}$
• D. $12\mu \text{m}$

Answer 1

The correct option is A $24\mu \text{m}$

Initially, ${\beta }_{\text{initial}}=\frac{\lambda D}{d_i}$
$\Rightarrow 0.25\times {10}^{-3}=\frac{\lambda }{d}\times 1$
$\Rightarrow \frac{\lambda }{d}=2.5\times {10}^{-4}......\left(1\right)$

Afterwards
$\frac{2}{3}{\beta }_i=\frac{\lambda D}{(d+\Delta d)}$
$\Rightarrow \frac{\lambda }{d+\Delta d}=\frac{2}{3}\times \frac{2.5\times {10}^{-4}}{1}.....\left(2\right)$

Dividing $\left(1\right)$ by $\left(2\right)$
$\frac{d+\Delta d}{d}=\frac{3}{2}\Rightarrow d=2\left(\Delta d\right)=2.4\text{mm}$
& $\lambda =2.4\times 2.5\times {10}^{-7}\text{m}$




Now, $P$ becomes central maxima.
$\Rightarrow$ for point $P$, $d\sin \theta =(\mu -1)t$
As $\theta$ is small, $\sin \theta \approx \tan \theta$

$\Rightarrow d\frac{20\beta }{D}=(\mu -1)t$

$\Rightarrow \frac{d}{D}\times \frac{20\times \lambda D}{d}=(1.5-1)t$

$\Rightarrow t=\frac{20\lambda }{0.5}=\frac{20\times 600\times {10}^{-9}}{0.5}=24\mu \text{m}.$

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