Question

Light takes time $t_1$ to travel a distance $x_1$ in vacuum and the same light takes time $t_2$ to travel a distance $x_2$ in a medium. The critical angle for that medium is

• A. ${sin}^{-1}\left(\frac{x_2{t}_2}{x_1{t}_1}\right)$
• B. ${sin}^{-1}\left(\frac{x_1{t}_2}{x_2{t}_1}\right)$
• C. ${sin}^{-1}\left(\frac{x_1{t}_1}{x_2{t}_2}\right)$
• D. ${sin}^{-1}\left(\frac{x_2{t}_1}{x_1{t}_2}\right)$

Answer 1

The correct option is B ${sin}^{-1}\left(\frac{x_1{t}_2}{x_2{t}_1}\right)$

$\begin{array}{c}\text{Velocity}{v}_1\text{in medium of}\\ \text{refractive index}{n}_1=1\text{(vacuum)}\end{array}$

$\begin{array}{c}{v}_1=\frac{x_1}{t_1},\text{similarly}{v}_2=\frac{x_2}{t_2}\\ \text{let}{n}_2=n\end{array}$

$\begin{array}{c}\text{we know}\\ \frac{v_1}{v_2}=\frac{n_2}{n_1}\\ \Rightarrow \frac{x_1}{x_2}\frac{t_2}{t_1}=n\end{array}$

$\begin{array}{c}\text{also, for critical angle}\\ \begin{array}{cc}{n}_1\sin i_c& =n_2\sin {90}^{\circ }\\ 1\times \sin i_c& =n\\ i_c& ={\sin }^{-1}\left(n\right)\\ i_c& ={\sin }^{-1}\left(\frac{x_1{t}_2}{x_2{t}_1}\right)\end{array}\end{array}$