Question

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

• A. $Sin^ {-1}(Tan \,i)$
• B. $Tan^{-1} (Sin\,i)$
• C. $Sin^{-1} (Cot\,i)$
• D. $Cos^{-1} (Tan\,i)$

Answer 1

Answer: (C)

$Sin^{-1} (Cot\,i)$

Answer 2

From law of reflection, $\angle i=\angle r\,\,\,\,\,\,\,\,\,\,\,...(i)$
and $\frac{\sin r^{\prime}}{\sin i}=\frac{\mu_{d}}{\mu_{r}}\,\,\,\,\,\,\,\,\,\,...(ii)$
$r+r'+90^{\circ}=180^{\circ}$
$\Rightarrow \,\,\,\, r+r^{\prime}=90^{\circ}$
or $\,\,\,\,\,i+r'=90^{\circ}$
$r'=\left(90^{\circ}-i\right)\,\,\,\,\,\,\,\,\,\,\,...(iii)$
From E (ii) $\frac{\sin \left(90^{\circ}-i\right)}{\sin i}=\frac{\mu_{d}}{\mu_{r}}$
or $\,\,\,\,\,\frac{\cos i}{\sin i}=\frac{\mu_{d}}{\mu_{r}} \Rightarrow \cot i=\frac{\mu_{d}}{\mu_{r}}$
But $\frac{\mu_{d}}{\mu_{r}}=\sin C$ (where $C$ is critical angle)
$\therefore \,\,\,\,\, \cot i=\sin C$
$\Rightarrow C=\sin ^{-1}(\cot i)$