Question

A ray of light from a denser medium strikes the plane boundary of a rarer medium at an angle of incidence $I$ . The angle of refraction is $r$ . If the reflected and the refracted rays are mutually perpendicular the critical angle of the medium is :

• A. $si{n}^{-1}\left(cotI\right)$
• B. $si{n}^{-1}\left(tanI\right)$
• C. $ta{n}^{-1}\left(sinI\right)$
• D. $si{n}^{-1}\left(tanr\right)$

Answer 1

Answer: (B)

$si{n}^{-1}\left(tanI\right)$

Angle between reflected ray and refracted ray is ${90}^o$

so according to diagram $r_1+r_2+{90}^o={180}^o⟹r_1+r_2={90}^o$

in question $r_2=r$

from law of reflection $i=r_1$, substituting values in above equation

$⟹i+r={90}^o$ where $r$ is angle of refraction.

From Snell's Law, $n_2\sin i=n_1\sin r$

$n.\sin i=\sin r$

$n.\sin i=\sin {90}^o-i$

$n.\sin i=\cos i$

$n=\cot i$

we know ${\theta }_{cric}={\sin }^{-1}\frac{1}{n}$

putting $n=\cot i$

${\theta }_{cric}={\sin }^{-1}\frac{1}{\cot i}$

${\theta }_{cric}={\sin }^{-1}\tan i$

Hence option B is correct