Question

In vacuum, in order to travel a distance $d$, the light may take time $t$ and in a medium to travel distances $5d$, it takes time $T$. Calculate the critical angle of the medium?
$\begin{aligned} & A.{{\sin }^{-1}}\left( \dfrac{5T}{t} \right) \\\ & B.{{\sin }^{-1}}\left( \dfrac{5t}{3T} \right) \\\ & C.{{\sin }^{-1}}\left( \dfrac{5t}{T} \right) \\\ & D.{{\sin }^{-1}}\left( \dfrac{3t}{5T} \right) \\\ \end{aligned}$

Answer 1

The critical angle of a system has been calculated by taking the inverse of the sine of the reciprocal of the refractive index of the material. Refractive index can be found by taking the ratio of the velocity of the light in vacuum to the velocity of the light in the medium. Velocity can be found by taking the ratio of the distance travelled to the time taken. This all will help you in answering this question.

Complete step by step answer:
The critical angle can be found by the use of equation given as,
$\theta ={{\sin }^{-1}}\left( \dfrac{1}{\mu } \right)$
Where $\mu$ be the refractive index of the material with respect to vacuum.
The refractive index can be found by taking the ratio of the velocity of light in vacuum to the velocity of the light in a specific medium. This can be expressed in an equation as,
$\mu =\dfrac{C}{V}$
Where $C$ be the velocity of light in vacuum and $V$ be the velocity of light in a medium.
Velocity can be found by taking the ratio of the distance travelled to the time taken.
Distance travelled in vacuum is given as $d$ and the time taken is $t$. The distance travelled in the medium is $5d$ and the time taken for this is given as $T$. That is,
$\begin{aligned} & C=\dfrac{d}{t} \\\ & V=\dfrac{5d}{T} \\\ \end{aligned}$
Substituting this value in the equation will give,
$\mu =\dfrac{\dfrac{d}{t}}{\dfrac{5d}{T}}=\dfrac{T}{5t}$
Substituting this value in the equation of critical angle will give,
$\theta ={{\sin }^{-1}}\left( \dfrac{5t}{T} \right)$
This has been given as option C.

Note:
The critical angle is defined as the specific angle of incidence where the ray of refraction is perpendicular to the normal. For this the light should travel from a denser medium to a rarer medium. The angle above the critical angle will cause the total internal reflection.