Question

The refractive index of water and glycerine are $1.33$ and $1.47$ respectively. What is the critical angle for a light ray going from the later to the former?

• A. $60{}^{\circ }48^{\prime }$
• B. $64{}^{\circ }48^{\prime }$
• C. $74{}^{\circ }48^{\prime }$
• D. None of these

Answer 1

Answer: (B)

When ray passes from denser to rarer medium, angle of refraction is greater than angle of incidence.
When a ray of light passes from glycerine (denser, $\mu =1.47$ ) to water (rarer, $\mu =1.33$ ) the angle of refraction (r) is greater than angle of incidence (i), than from Snells law
$\frac{\sin i}{\sin r}=\frac{{\mu }_2}{{\mu }_1}<1$
When $r=90^{\circ }$, corresponding angle of incidence is known as critical angle i.e., $i={\theta }_c$.
$\therefore \frac{\sin {\theta }_c}{\sin 90^{\circ }}=\frac{{\mu }_2}{{\mu }_1}$
$\Rightarrow \sin {\theta }_c=\frac{{\mu }_2}{{\mu }_1}$
$\Rightarrow {\theta }_c={\sin }^{-1}\left(\frac{{\mu }_2}{{\mu }_1}\right)$
$={\sin }^{-1}\left(\frac{1.33}{1.47}\right)$
${\theta }_c=64^{\circ }48$