Question

Light takes $T_1$ sec to travel a distance $x$ In a vacuum and the same light takes $T_2$ sec to travel $10cm$ in a medium. The critical angle for the corresponding medium will be:

Answer 1

Critical angle:

When light ray traveling in the denser medium is incident on the rarer medium at an angle of incidence such that the refracted ray grazes out through the interface boundary (∠r=90) then the angle of incidence is called the critical angle.

Step 1. Given data:

In the case of vacuum, Time=$T_1$, Distance=$x$,

In the case of a medium, Time= $T_2$, Distance= $10cm$

Step 2. Formula used:

$\sin \left(i_c\right)=\frac{v}{c}$

Where $i_c$=critical angle, $c$= Velocity of light in vacuum, $v$= Velocity of light in a medium.

Step 3. Explanation:

$c$= Velocity of light in vacuum.

$v$= Velocity of light in a medium.

Here, we can clearly see, $c=\frac{x}{T_1},v=\frac{10}{T_2}$

Where $c,x,v,T_{1}and{T}_2$ have their usual meanings.

Also, $$\begin{gathered} \sin \left(i_c\right)=\frac{1}{\mu }=\frac{v}{c}=\frac{10}{T_2}\times \frac{T_1}{x}=\left(\frac{10.T_1}{T_2.x}\right) \\ {i}_c={\sin }^{-1}\left(\frac{10.T_1}{T_2.x}\right) \end{gathered}$$

Hence, the critical angle for the corresponding medium will be, $i_c={\sin }^{-1}\left(\frac{10T_1}{T_{2}x}\right)$.