A, B and C are the parallel sided transparent media of refractive indices \(\mathrm{n}_{1}, \mathrm{n}_{2}\) and \(\mathrm{n}_{3}\) 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 madia B and C. Then, \(\sin i\) equals to

Applying Snell's law between the surfaces \(\mathrm{A}\) and \(\mathrm{B}\)



\[
\mathrm{n}_{1} \sin \mathrm{i}=\mathrm{n}_{2} \sin \mathrm{r}_{1} \quad \text{...(i)}
\]
Again applying Snell's law between surfaces \(B\) and C
\(\mathrm{n}_{2} \sin \mathrm{r}_{1}=\mathrm{n}_{3} \sin \mathrm{r}_{2} \quad \text{...(ii)}\)
From Eqs. (i) and (ii), we get
\[
\begin{aligned}
\text { Here, } & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \sin \mathrm{r}_{2} \\
\therefore & \mathrm{r}_{2} &=90^{\circ} \\
\Rightarrow & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \\
\Rightarrow & & \sin \mathrm{i} &=\frac{\mathrm{n}_{3}}{\mathrm{n}_{1}}
\end{aligned}
\]