JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
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
- A \(\frac{1}{{{{\cos }^{ - 1}}\,\left( {\sin {\mkern 1mu} {\theta _C}} \right)}}\)
- B \(\frac{1}{{{{\tan }^{ - 1}}\,\left( {\sin {\mkern 1mu} {\theta _C}} \right)}}\)
- C \({{{\cos }^{ - 1}}\left( {\sin \,{\theta _C}} \right)}\)
- D \({{{\tan }^{ - 1}}\left( {\sin \,{\theta _C}} \right)}\)
Answer & Solution
Correct Answer
(D) \({{{\tan }^{ - 1}}\left( {\sin \,{\theta _C}} \right)}\)
Step-by-step Solution
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,…
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