Question 66 of 295

Let the refractive index of a denser medium with respect to 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${\tan }^{-1}⁡\left(\sin ⁡{\theta }_{C}\right)$B$\dfrac{1}{{\tan }^{-1}⁡\left(\sin ⁡{\theta }_{C}\right)}$C${\cos }^{-1}⁡\left(\sin ⁡{\theta }_{C}\right)$D$\dfrac{1}{{\cos }^{-1}⁡\left(\sin ⁡{\theta }_{C}\right)}$

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Answer: (A) ${\tan }^{-1}⁡\left(\sin ⁡{\theta }_{C}\right)$

In accordance with Snell's law, $\mu =\dfrac{{\mu }_{R}}{{\mu }_{D}}=\dfrac{\sin ⁡{i}_{c}}{\sin ⁡{90}^{o}}$
here ${i}_{c}$ is the angle of incidence.

$\dfrac{{\mu }_{R}}{{\mu }_{D}}=\sin ⁡{i}_{c}$

$\mu =\dfrac{{\mu }_{R}}{{\mu }_{D}}=\dfrac{\sin ⁡A}{\sin ⁡r}$

$=\dfrac{\sin A}{\sin \left(90^{\circ}-A\right)}=\dfrac{\sin A}{\cos A}$

$\dfrac{{\mu }_{R}}{{\mu }_{D}}=\tan ⁡A$.

The critical angle is related to the angle $A$ as $\tan ⁡A=\sin ⁡{\theta }_{C}$

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