Q. If a ray of light in a denser medium enters into a rarer medium at an angle of incidence $i,$ the angle of reflection and reflection are respectively $r\ and\ r'.$If the reflected and refracted rays are at right angles to each other, the critical angle for the given pair of media is
A) $\sin^{- 1}{(\tan{r'})}$
B) $\sin^{- 1}{(\tan r)}$
C) ${\tan^{- 1}(}{\sin{i)}}$
D) $\cot{(\tan i)}$
Explanation
(b) Here $i = r$
$r'' = 90{^\circ} - r$
So, $\mu = \frac{\sin{r''}}{\sin r} = \frac{sin(90{^\circ} - r)}{\sin r}$
$\mu = \frac{\cos r}{\sin r} = \frac{1}{\tan r}$
But $\mu = \frac{1}{\sin C}$
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Where $C$ is the critical angle.
So, $\frac{1}{\sin C} = \frac{1}{\tan r}$
$\Rightarrow \sin{C = \tan r}$
$Or\ \ C = \sin^{- 1}{(\tan r)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.