Question

A ray of light passes through four transparent media with refractive indices $\mu_{1}, \mu_{2}, \mu_{3}$ and $\mu_{4}$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$, we must have :

• A. $ {{\mu }_{1}}={{\mu }_{2}} $
• B. $ {{\mu }_{2}}={{\mu }_{3}} $
• C. $ {{\mu }_{3}}={{\mu }_{4}} $
• D. $ {{\mu }_{1}}={{\mu }_{4}} $

Answer 1

Answer: (D)

According to given figure
${ }_{1} \mu_{2} =\frac{\sin i}{\sin r_{1}} \ldots$ (1)
${ }_{2} \mu_{3} =\frac{\sin r_{1}}{\sin r_{2}} \ldots$(2)
and $\mu_{4} =\frac{\sin r_{2}}{\sin r_{3}}\ldots$ (3)
On multiplying eqs. $(1),(2)$ and $(3)$, we get
$\frac{\sin i}{\sin r_{1}} \times \frac{\sin r_{1}}{\sin r_{2}} \times \frac{\sin r_{2}}{\sin r_{3}}$
$=\frac{\mu_{2}}{\mu_{1}} \times \frac{\mu_{3}}{\mu_{2}} \times \frac{\mu_{4}}{\mu_{3}}$
or $ \frac{\sin i}{\sin r_{3}}=\frac{\mu_{4}}{\mu_{1}} $
or $ \mu_{1} \sin i=\mu_{4} \sin r_{3}$
Since, $A B$ is parallel to $C D$ so, $i=r_{3}$
Therefore, $\mu_{1}=\mu_{4}$