Q. A ray of light from a denser medium strikes a rarer medium at angle of incidence $i$. The reflected and refracted rays make an angle of $90{^\circ}$ with each other. The angles of reflection and refraction are $r$ and $r’$ respectively. The critical angle is
A) $\sin^{- 1}{(\tan{r')}}$
B) $\sin^{- 1}{(\tan{r)}}$
C) $\tan^{- 1}{(\tan{r')}}$
D) $\tan^{- 1}{(\tan{i)}}$
Explanation
(b) [image:9_image103.emf]
$\sin{C = \frac{1}{\mu} = \frac{1}{\sin{r^{''}/\sin i}} = \frac{\sin i}{\sin{r''}}}$
As is clear as shown in figure
$\angle CBD = 90{^\circ}$
$\therefore 90{^\circ} - r + 90{^\circ} - r^{''} = 90{^\circ}$
Or ${\ r}^{''} = 90{^\circ} - r$
$\therefore\sin{C = \frac{\sin i}{\sin{(90{^\circ} - r)}} = \frac{\sin i}{\cos r}}$
$= \frac{\sin i}{\cos i} = \tan{i\ (\because i = r)}$
$C = \sin^{- 1}{(\tan{i) = \sin\ ^{- 1}(\tan{r)}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.