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A ray of light enters from a rarer to a denser medium. The angle of incidence is Then the reflected and refracted rays are mutually perpendicular to each other. The critical angle for the pair of media is
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Verified Answer
The correct answer is:
$\sin ^{-1}(\cot \mathrm{i})$
From law of reflection, $\angle \mathrm{i}=\angle \mathrm{r}$ and
$$
\frac{\sin \mathrm{r}^{\prime}}{\sin \mathrm{i}}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}
$$
From the figure
$$
r^{\prime}=\left(90^{\circ}-i\right)
$$
From Eq. (ii) $\frac{\sin \left(90^{\circ}-i\right)}{\sin i}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}$ or $\quad \frac{\cos \mathrm{i}}{\sin \mathrm{i}}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}} \Rightarrow \cot \mathrm{i}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}$
But $\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}=\sin \mathrm{C}$ (where is critical angle) $\therefore \quad \cot \mathrm{i}=\sin \mathrm{C} \Rightarrow \mathrm{C}=\sin ^{-1}(\cot \mathrm{i})$
$$
\frac{\sin \mathrm{r}^{\prime}}{\sin \mathrm{i}}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}
$$
From the figure
$$
r^{\prime}=\left(90^{\circ}-i\right)
$$
From Eq. (ii) $\frac{\sin \left(90^{\circ}-i\right)}{\sin i}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}$ or $\quad \frac{\cos \mathrm{i}}{\sin \mathrm{i}}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}} \Rightarrow \cot \mathrm{i}=\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}$
But $\frac{\mu_{\mathrm{d}}}{\mu_{\mathrm{r}}}=\sin \mathrm{C}$ (where is critical angle) $\therefore \quad \cot \mathrm{i}=\sin \mathrm{C} \Rightarrow \mathrm{C}=\sin ^{-1}(\cot \mathrm{i})$
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