Question

The speed of light in media $M_1$ and $M_2$ are $1.5\times {10}^{8}m{s}^{-1}$ and $2\times {10}^{8}m{s}^{-1}$ respectively. A ray travels from medium $M_1$ to the medium $M_2$ with an angle of incidence $\theta$. The ray suffers total internal reflection. Then the value of the angle of incidence $\theta$ is.

• A. $>{\sin }^{-1}\left(\frac{3}{4}\right)$
• B. $<{\sin }^{-1}\left(\frac{3}{4}\right)$
• C. $={\sin }^{-1}\left(\frac{4}{3}\right)$
• D. $\le {\sin }^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is A $>{\sin }^{-1}\left(\frac{3}{4}\right)$

Velocity in (denser) medium $M_1$ $v_1=1.5\times {10}^8$ $m/s$

Velocity in (rarer) medium $M_2$ $v_2=2\times {10}^8$ $m/s$

For total internal reflection to occur, the angle of incidence $\left(\theta \right)$ must be greater than critical angle ${\theta }_c$.

$\therefore$ $\theta >{\sin }^{-1}(\frac{n_r}{n_d})$

Using $\frac{n_r}{n_d}=\frac{v_1}{v_2}$ we get $\theta >{\sin }^{-1}(\frac{v_1}{v_2})$

$\therefore$ $\theta >{\sin }^{-1}(\frac{1.5\times {10}^8}{2\times {10}^8})$ $⟹$ $\theta >{\sin }^{-1}(\frac{3}{4})$