Question

The critical angle for refraction from medium $1$ to air is ${\theta }_1$ and that from medium $2$ to air is ${\theta }_2$. The critical angle for refraction from medium $2$ to medium $1$ will be

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

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{\sin {\theta }_2}{\sin {\theta }_1}\right)$

For medium $1$ and air,
${\mu }_1\sin i_c={\mu }_a\sin {90}^{\circ }$
$\Rightarrow {\mu }_1\sin {\theta }_1=1$
$\Rightarrow {\mu }_1=\frac{1}{\sin {\theta }_1}$ ...........$\left(i\right)$
For medium $2$ and air,
${\mu }_2\sin i_c={\mu }_a\sin {90}^{\circ }$
$\Rightarrow {\mu }_2\sin {\theta }_2=1$
$\Rightarrow {\mu }_2=\frac{1}{\sin {\theta }_2}$ ...........$\left(ii\right)$
Now for medium $2$ and medium $1$,
${\mu }_2\sin i_c={\mu }_1\sin {90}^{\circ }$
$\Rightarrow {\mu }_2\sin i_c={\mu }_1$
$\Rightarrow \frac{\sin i_c}{\sin {\theta }_2}=\frac{1}{\sin {\theta }_1}$
$[$from $\left(i\right)$ and $\left(ii\right)]$
$\Rightarrow \sin i_c=\frac{\sin {\theta }_2}{\sin {\theta }_1}$
$\Rightarrow i_c={\sin }^{-1}\left(\frac{\sin {\theta }_2}{\sin {\theta }_1}\right)$

Why this question? Tips: Instead of going for formula based approach, it is recommended to apply Snell's law for each pair of media and then couple it with condition of critical incidence.