Question

A ray of light passes through $4$ transparent media with refractive indices ${\mu }_1,{\mu }_2,{\mu }_3$ and ${\mu }_4$ as shown. The surface of all media are parallel. If the emergent ray $\text{CD}$ is parallel to the incident ray $\text{AB}$, 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

The correct option is D ${\mu }_1={\mu }_4$

Here, incident ray $\text{AB}$ is parallel to emergent ray $\text{CD}$

$\therefore {\theta }_1={\theta }_4$

Since, incident ray $\text{AB}$ is parallel to emergent ray $\text{CD}$, so we can use

$\mu \sin \theta =\text{constant}$

So, for first and last media,

${\mu }_1\sin {\theta }_1={\mu }_4\sin {\theta }_4$

$\Rightarrow {\mu }_1={\mu }_4(\because {\theta }_1={\theta }_4)$

Hence, option $\left(d\right)$ is correct.