Q.
$A,\ B\ and\ C$are the parallel sided transparent media of refractive indices $n_{1},\ n_{2}andn_{3}$ respectively. They are arranged as shown in the figure. A ray is incident at an angle $i$ on the surface of separation of $A\ and\ B$ which is as shown in the figure. After the refraction into the medium $B,$ the ray grazes the surface of separation of the media $B$ and $C.$ Then, $\sin i$ equal to
A) $\frac{n_{3}}{n_{1}}$
B) $\frac{n_{1}}{n_{3}}$
C) $\frac{n_{2}}{n_{3}}$
D) $\frac{n_{1}}{n_{2}}$
Explanation
(a) Applying Snell’s law between the surfaces $A$ and $B$
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$n_{1}\sin i = n_{2}\sin r_{1}\ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Again applying Snell’s law between surfaces $B$ and $C$
$n_{2}\sin r_{1} = n_{3}\sin r_{2}\ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we get
$n_{1}isn\ i = \ n_{3}\sin r_{2}$
Here, $r_{2}\ = 90{^\circ}$
$\therefore n_{1}\sin i = n_{3}$
$\Rightarrow \sin{i = \frac{n_{3}}{n_{1}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.