Question

A Young's double- slit apparatus is immersed in a liquid of refractive index $1.33$. It has a slit separation of $1\text{mm}$ and an interference pattern is observed on screen at a distance $1.33\text{m}$ from the plane of slits. The wavelength of light in air is $6300\dot{\text{A}}$. If one of the slits is covered by a glass sheet of $\mu =1.53$, find the smallest thickness of sheet to interchange the position of maxima and minima.

• A. $2.57\mu \text{m}$
• B. $1.57\mu \text{m}$
• C. $3.27\mu \text{m}$
• D. $4.18\mu \text{m}$

Answer 1

The correct option is B $1.57\mu \text{m}$

In order to obtain maxima, the path differences between interfering waves is,

$\Delta x=n\lambda$

Similarly, for obtaining minima,

$\Delta x=(2n-1)\frac{\lambda }{2}$

We know that if path difference between waves is altered by $\frac{\lambda }{2}$ then position of maxima and minima will interchange.

By introducing glass sheet path difference created.

$\Delta x=({\mu }_{rel}-1)t$

According to question,
$({\mu }_{rel}-1)t=\frac{{\lambda }_w}{2}$

$\Rightarrow (\frac{{\mu }_2}{{\mu }_1}-1)t=\frac{{\lambda }_w}{2}$

$\Rightarrow (\frac{1.53}{1.33}-1)t=\frac{\frac{(6300\times {10}^{-10})}{1.33}}{2}[\because {\lambda }_w=\frac{{\lambda }_{air}}{{\mu }_w}]$

$\Rightarrow \frac{0.20}{1.33}t=2368.4\times {10}^{-10}$

$\Rightarrow t=1.574\times {10}^{-6}\text{m}$

$\therefore t_{min}=1.57\mu \text{m}$

Hence, $\left(\text{B}\right)$ is the correct answer.