Optics Ques 31
- A ray of light from a denser medium strikes a rarer medium at an angle of incidence $i$ (see figure). The reflected and refracted rays make an angle of $90^{\circ}$ with each other. The angles of reflection and refraction are $r$ and $r^{\prime}$. The critical angle is $(1983,1 M)$
(a) $\sin ^{-1}(\tan r)$
(b) $\sin ^{-1}(\cot i)$
(c) $\sin ^{-1}\left(\tan r^{\prime}\right)$
(d) $\tan ^{-1}(\sin i)$
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Answer:
Correct Answer: 31.(a)
Solution:
Formula:
Critical Angle and Total Internal Reflection (T. I. R.)
- $r+r^{\prime}+90^{\circ}=180^{\circ}$
$$ \begin{aligned} & \therefore & r^{\prime} & =90^{\circ}-r \\ & \text { Further, } & i & =r \end{aligned} $$
Applying Snell’s law, $\mu _D \sin i=\mu _R \sin r^{\prime}$
or $\mu _D \sin r=\mu _R \sin \left(90^{\circ}-r\right)=\mu _R \cos r$
$\therefore \frac{\mu _R}{\mu _D}=\tan r, \theta _C=\sin ^{-1} \frac{\mu _R}{\mu _D}=\sin ^{-1}(\tan r)$
BETA
