Question

In vacuum, to travel distance 'd', light takes time 't' and in medium to travel distance '5d', it takes time 'T'. The critical angle of the medium is.
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. ${\sin ^{ - 1}}\left( {\dfrac{{3t}}{{5t}}} \right)$

Answer 1

We know that the sine of the medium's critical angle is inversely proportional to the refractive index of that medium. We also know that the medium's refractive index is given as the ratio speed of light in vacuum to the medium's speed.

Complete step by step answer:
Given:
The travel distance of light in the vacuum is ${d_1} = d$.
The time taken by light to travel in the vacuum is ${t_1} = t$.
The travel distance of light in a medium is ${d_2} = 5d$.
The time taken by light to travel in the medium is ${t_2} = T$.
We have to find the expression for the critical angle of the medium.
Let us write the expression for the refractive index of the given medium with respect to vacuum.
$\mu = \dfrac{c}{v}$……(1)
Here c is the speed of light in vacuum, v is the speed of light in the medium.
We can write the expression for the speed of light in vacuum as below:
$c = \dfrac{{{d_1}}}{{{t_1}}}$
On substituting d for ${d_1}$ and t for ${t_1}$ in the above expression, we get:
$c = \dfrac{d}{t}$
The expression for the speed of light in the medium can be written as below:
$v = \dfrac{{{d_2}}}{{{t_2}}}$
On substituting 5d for ${d_1}$ and T for ${t_1}$ in the above expression, we get:
$v = \dfrac{{5d}}{T}$
Substitute $\dfrac{d}{t}$ for c and $\dfrac{{5d}}{T}$ for v in equation (1).

\mu = \dfrac{{\left( {\dfrac{d}{t}} \right)}}{{\left( {\dfrac{{5d}}{T}} \right)}}\\\ = \dfrac{T}{{5t}} \end{array}$$ We know that the expression for critical angle can be written as below: $$\sin {i_c} = \dfrac{1}{\mu }$$ Here $${i_c}$$ is the critical angle of the medium. On substituting $$\dfrac{T}{{5t}}$$ for $$\mu $$ in the above expression, we get: $$\begin{array}{l} \sin {i_c} = \dfrac{1}{{\left( {\dfrac{T}{{5t}}} \right)}}\\\ \sin {i_c} = \dfrac{{5t}}{T} \end{array}$$ Taking the inverse of the above equation, we get: $${i_c} = {\sin ^{ - 1}}\left( {\dfrac{{5t}}{T}} \right)$$ Therefore, the expression for the medium's critical angle can be written as $${\sin ^{ - 1}}\left( {\dfrac{{5t}}{T}} \right)$$, and option (C) is correct. **Note:** We can consider the angle of incidence as a critical angle when the angle of refraction becomes a right angle. This phenomenon can happen when light travels from a denser medium to a lighter medium.