Question

$A, B$ and $C$ are the parallel sided transparent media of refractive index $n_1, n_2$ and $ n_3$ respectively. They are arranged as shown in the figure. A ray is incident at an angle $\theta$ 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, $Sin\, \theta$ = .........

• A. $\frac {n_3}{n_1}$
• B. $\frac {n_1}{n_2}$
• C. $\frac {n_2}{n_3}$
• D. $\frac {n_1}{n_3}$

Answer 1

Answer: (A)

Applying Snell's law between the surfaces A and B

$n_{1} \sin i=n_{2} \sin r_{1}$...(i)
Again applying Snell's law between surfaces $\underline{B}$ and $C$
$n_{2} \sin r_{1}=n_{3} \sin r_{2}$...(ii)
From Eqs. (i) and (ii), we get
$n_{1} \sin i=n_{3} \sin r_{2}$
Here, $r_{2}=90^{\circ}$
$\therefore n_{1} \sin i =n_{3}$
$\Rightarrow \sin i =\frac{n_{3}}{n_{1}}$