Question

The figure below shows the paths of a portion of a ray of light as it passes through three different materials. Mark the correct relation among refractive indices of three materials :

• A. ${\mu }_1>{\mu }_2>{\mu }_3$
• B. ${\mu }_2>{\mu }_1>{\mu }_3$
• C. ${\mu }_1>{\mu }_3>{\mu }_2$
• D. ${\mu }_1<{\mu }_3<{\mu }_2$

Answer 1

The correct option is D ${\mu }_1<{\mu }_3<{\mu }_2$

From the figure,

${\theta }_1>{\theta }_3>{\theta }_2$

Applying snells law at $P-1$ and $P-2$,

${\mu }_1\sin \left({\theta }_1\right)={\mu }_2\sin \left({\theta }_2\right)$ ………(i)

${\mu }_2\sin \left({\theta }_2\right)={\mu }_3\sin \left({\theta }_3\right)$ …….(ii)

From (i) and (ii)

${\mu }_1\sin \left({\theta }_1\right)={\mu }_2\sin \left({\theta }_2\right)={\mu }_3\sin \left({\theta }_3\right)=k$ …….(iii)

As ${\theta }_1>{\theta }_3>{\theta }_2$

$\Rightarrow \sin \left({\theta }_1\right)>\sin \left({\theta }_3\right)>\sin \left({\theta }_2\right)$

From (iii),

${\mu }_1=\frac{k}{\sin {\theta }_1},{\mu }_2=\frac{k}{\sin {\theta }_2},{\mu }_3=\frac{k}{\sin \left({\theta }_3\right)}$

So,

As $\sin \left({\theta }_1\right)>\sin \left({\theta }_3\right)>\sin \left({\theta }_2\right)$,

$\Rightarrow {\mu }_1<{\mu }_3<{\mu }_2$.