Question

In young's double slit experiment, the fringes are displaced by a distance $x$ when a glass plate of refractive index 1.5 is introduced in the path of one of the beams. When this plate is replaced by another plate of the same thickness, the shift of the fringes is ${\displaystyle \frac{3}{2}}x$. The refractive index of the second plate is:
A. 1.75
B. 1.50
C. 1.25
D. 1.00

Answer 1

Hint: This question is from the topic Wave Optics. We have to consider the Young’s double slit experiment. And we have to use the results of the condition in which a glass plate is inserted in the path of the beams.

Complete step by step answer:
When a plate made of an optical medium is inserted in the path of the rays of Young’s Double Slit Experiment, the fringes which are formed are slightly displaced. This is known as the shifting of the fringes.
The shift in the Fringes when a plate is introduced is given by:
$x={\displaystyle \frac{(\mu -1)t\beta }{\lambda }}$, where $\mu$ is the refractive index of the plate, $t$ is the thickness of the plate, $\beta$ is the fringe width and $\lambda$ is the wavelength of the light.

If we first consider the case when plate of refractive index 1.5 is introduced then what will be the shift.:
$$\begin{gathered} x={\displaystyle \frac{(\mu -1)t\beta }{\lambda }} \\ ⟹x={\displaystyle \frac{(1.5-1)t\beta }{\lambda }} \end{gathered}$$
Now we will try to find the case when another plate of unknown refractive index is introduced in the path.
${\displaystyle \frac{3}{2}}x={\displaystyle \frac{({\mu }_2-1)t\beta }{\lambda }}$
Here we have three quantities whose value is not given to us and also not required; Therefore, we will eliminate those quantities.
We will divide the equation of shift in fringe width of both the cases:
$$\begin{gathered} {\displaystyle \frac{x}{{\displaystyle \frac{3}{2}}x}}={\displaystyle \frac{{\displaystyle \frac{(1.5-1)t\beta }{\lambda }}}{{\displaystyle \frac{({\mu }_2-1)t\beta }{\lambda }}}} \\ {\displaystyle \frac{2}{3}}={\displaystyle \frac{(1.5-1)}{({\mu }_2-1)}} \\ ⟹2{\mu }_2-2=3\times 0.5 \\ ⟹2{\mu }_2=3.5 \\ ⟹{\mu }_2=1.75 \end{gathered}$$

Hence, we have calculated the refractive index of the second plate and it is 1.75, and the correct option is option (A).

Note:
These kinds of questions require deep knowledge of the Young’s Double slit experiment. The expression of the shift of fringe width is very important to solve this type of question. Also, students should be aware of the fact about the relative refractive indices to avoid confusion.