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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
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Verified Answer
The correct answer is:
$\frac{\mathrm{n}_{3}}{\mathrm{n}_{1}}$
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
$$
\begin{aligned}
\text { Here, } & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \sin \mathrm{r}_{2} \\
\therefore & \mathrm{r}_{2} &=90^{\circ} \\
\Rightarrow & \mathrm{n}_{1} \sin \mathrm{i} &=\mathrm{n}_{3} \\
\Rightarrow & & \sin \mathrm{i} &=\frac{\mathrm{n}_{3}}{\mathrm{n}_{1}}
\end{aligned}
$$
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