Let the refractive index of a denser medium with respect to a rarer medium be \(n_{12}\) and its critical angle be \(\theta_C\). At an angle of incidence \(A\) when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is \(90^o\). Angle \(A\) is given by

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Answer & Solution

Correct Answer

(D) \({{{\tan }^{ - 1}}\left( {\sin \,{\theta _C}} \right)}\)

Detailed explanation

From Snell's law, \(\frac{\mu_{\mathrm{R}}}{\mu_{\mathrm{D}}}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) ..... \((i)\)

\(\because \,\angle {\text{i}} = {\text{A}}\) and \(\angle {\text{r}} = \left( {{{90}^o} - {\text{A}}} \right)\)

We also know that, \(\sin \theta_{C}=\frac{\mu_{R}}{\mu_{D}}\)

From \(eq^{n}\,\,(i)\), \(\sin {\theta _C} = \frac{{\sin A}}{{\sin \left( {{{90}^o} - A} \right)}}\)

\(\sin {\theta _{\text{C}}} = \frac{{\sin {\text{A}}}}{{\cos {\text{A}}}}\)

\(\sin {\theta _{\text{C}}} = \tan {\text{A}}\)

\({\text{A}} = {\tan ^{ - 1}}\left( {\sin {\theta _{\text{C}}}} \right)\)

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\((B)\) Galvanometer	\((II)\) Converts mechanical energy into electrical energy
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