Question

In vacuum, to travel distance 'd', light takes time 't' and in a medium to travel distances '$5d$', it takes time 'T'. The critical angle of the medium is

• A. ${\sin }^{-1}\left(\frac{5T}{t}\right)$
• B. ${\sin }^{-1}\left(\frac{5t}{3T}\right)$
• C. ${\sin }^{-1}\left(\frac{5t}{T}\right)$
• D. ${\sin }^{-1}\left(\frac{3t}{5T}\right)$

Answer 1

Answer: (C)

${\sin }^{-1}\left(\frac{5t}{T}\right)$

Definition of critical angle is: $i_c={\sin }^{-1}\left(\frac{1}{\mu }\right)$

Where, $\mu$ is the refractive index of the medium w.r.t. vacuum.

Definition of refractive index is:

$\mu =\frac{c}{v}=\frac{d/t}{5d/T}=\frac{T}{5t}$

Where, $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium.

So, $i_c={\sin }^{-1}\left(\frac{5t}{T}\right)$