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 $CD$ is parallel to the incident ray $AB$, 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

Answer: (D)

According to given figure
${}_1{\mu }_2=\frac{\sin i}{\sin r_1}\dots$ (1)
${}_2{\mu }_3=\frac{\sin r_1}{\sin r_2}\dots$(2)
and ${\mu }_4=\frac{\sin r_2}{\sin r_3}\dots$ (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, $AB$ is parallel to $CD$ so, $i=r_3$
Therefore, ${\mu }_1={\mu }_4$