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A ray of light passes from a denser medium to a rarer medium at an angle of incidence $i$ . The reflected and refracted rays make an angle of $90^{\circ}$ with each other. The angle of reflection and refraction are resepectively $r$ and $r^{\prime}$ . The critical angle is given by:
Option: 1 $\sin ^{-1} (cotr)$

Option: 2 $\tan ^{-1}(\sin i)$

Option: 3 $\sin ^{-1}(\tan r')$
Option: 4 $\sin ^{-1}\: (tanr)$

Answers (1)

best_answer

$\begin{gathered} i=r \rightarrow \text { (i) (Law of refletion) } \\ \end{gathered}$

$\begin{gathered} r+r^{'}+90^{\circ}=180^{\circ} \\ \end{gathered}$

$\begin{gathered} {r'=90-r} \end{gathered}$

$\begin{gathered} {\mu_{1} \sin i=\mu_{2} \sin r^{\prime}} \\ \end{gathered}$

$\begin{gathered} \mu_{1} \sin (r)=\mu_{2} \sin (90-r) \\ \end{gathered}$

$\begin{gathered} \tan r=\frac{\mu_{2}}{\mu_{1}}=\frac{1}{\mu_{1}^{2}}=\sin i_{c} \\ \end{gathered}$

$\left [ \because \sin i_{c}=\frac{1}{\mu ^{rarer}_{denser}} \right ]$

$\sin i_{c}=\tan r \\$

$i_{c}=\sin ^{-1}(\tan r)$

The correct option is (4)

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vishal kumar

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