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A ray of light enters from a rarer to a denser medium. The angle of incidence is $i$. Then, the reflected and refracted rays are mutually perpendicular to each other. The critical angle for the pair of media is
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The correct answer is:
$\sin ^{-1}(\cot i)$
The angle of incidence $=i$
Let $r$ be the angle of refraction.
According to given situation, reflected rays and refracted rays are perpendicular to each other.
i.e. $\quad i+r=90^{\circ}$ $\therefore$ According to Snell's law,
$$
\begin{aligned}
\mu &=\frac{\sin i}{\sin r} \\
&=\frac{\sin i}{\sin (90-i)} \quad[\because \text { From Eq. (i) }] \\
&=\frac{\sin i}{\cos i} \\
\mu &=\tan i
\end{aligned}
$$
If $i_{C}$ be the critical angle for the pair of media, then
$$
\mu=\frac{1}{\sin i_{C}}
$$
From Eas. (ii) and (iii), we get
$\tan i=\frac{1}{\sin i_{C}}$
$\Rightarrow \quad \sin i_{C}=\cot i$
$\Rightarrow \quad i_{C}=\sin ^{-1}(\cot i)$
Let $r$ be the angle of refraction.
According to given situation, reflected rays and refracted rays are perpendicular to each other.
i.e. $\quad i+r=90^{\circ}$ $\therefore$ According to Snell's law,
$$
\begin{aligned}
\mu &=\frac{\sin i}{\sin r} \\
&=\frac{\sin i}{\sin (90-i)} \quad[\because \text { From Eq. (i) }] \\
&=\frac{\sin i}{\cos i} \\
\mu &=\tan i
\end{aligned}
$$
If $i_{C}$ be the critical angle for the pair of media, then
$$
\mu=\frac{1}{\sin i_{C}}
$$
From Eas. (ii) and (iii), we get
$\tan i=\frac{1}{\sin i_{C}}$
$\Rightarrow \quad \sin i_{C}=\cot i$
$\Rightarrow \quad i_{C}=\sin ^{-1}(\cot i)$
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