Q127 / 180 · KCET 2010 · Physics
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
- Correct answerA$\frac{\mathrm{n}_{3}}{\mathrm{n}_{1}}$
- B$\frac{\mathrm{n}_{1}}{\mathrm{n}_{3}}$
- C$\frac{\mathrm{n}_{2}}{\mathrm{n}_{3}}$
- D$\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}$
Solution
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 $$ \begin{aligned} \text { Here, } & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \sin \mathrm{r}_{2} \\ \therefore & \mathrm{r}_{2} &=90^{\circ} \\ \Rightarrow & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \\ \Rightarrow & & \sin \mathrm{i} &=\frac{\mathrm{n}_{3}}{\mathrm{n}_{1}} \end{aligned} $$
$$ \mathrm{n}_{1} \sin \mathrm{i}=\mathrm{n}_{2} \sin \mathrm{r}_{1} \quad \text{...(i)} $$ 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 $$ \begin{aligned} \text { Here, } & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \sin \mathrm{r}_{2} \\ \therefore & \mathrm{r}_{2} &=90^{\circ} \\ \Rightarrow & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \\ \Rightarrow & & \sin \mathrm{i} &=\frac{\mathrm{n}_{3}}{\mathrm{n}_{1}} \end{aligned} $$
Hence, the correct answer is option: 'A'.