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A ray of light travels from a denser medium to a rarer medium. The reflected and the refracted rays are perpendicular to each other. If ' $\mathrm{r}$ ' and ' $\mathrm{r}_1$ ' are the angle of reflection and refraction respectively and ' $\mathrm{C}$ ' is the critical angle, then the angle of incidence is
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Verified Answer
The correct answer is:
$\tan ^{-1}(\sin C)$
Since reflected and refracted rays are perpendicular to each other,
$$
\begin{aligned}
& r+r_1=90^{\circ} \\
& \therefore r_1=90^{\circ}-r
\end{aligned}
$$
Refractive index of the denser medium
$$
\begin{aligned}
& \mu=\frac{\sin r_1}{\sin r}=\frac{\sin \left(90^{\circ}-r\right)}{\sin r}=\frac{\cos r}{\sin r}=\cot r \\
& \text { Also, } \mu=\frac{1}{\sin C} \quad \therefore \frac{1}{\sin C}=\cot r \\
& \text { or } \sin C=\tan r=\tan i \\
& \therefore i=\tan ^{-1}(\sin C)
\end{aligned}
$$
$$
\begin{aligned}
& r+r_1=90^{\circ} \\
& \therefore r_1=90^{\circ}-r
\end{aligned}
$$
Refractive index of the denser medium
$$
\begin{aligned}
& \mu=\frac{\sin r_1}{\sin r}=\frac{\sin \left(90^{\circ}-r\right)}{\sin r}=\frac{\cos r}{\sin r}=\cot r \\
& \text { Also, } \mu=\frac{1}{\sin C} \quad \therefore \frac{1}{\sin C}=\cot r \\
& \text { or } \sin C=\tan r=\tan i \\
& \therefore i=\tan ^{-1}(\sin C)
\end{aligned}
$$
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