Question

A ray of light is incident at ${60}^{\circ }$ on a glass-air interface. What will happen to the ray?

${\mu }_g$ = $\sqrt{2}$, ${\mu }_a$ = 1

• A. The light ray will undergo total internal reflection.
• B. The light ray will get reflected making an angle of ${60}^{\circ }$ with the normal.
• C. The light ray will get refracted and bends away from the normal.
• D. The light ray will get refracted and bend towards the normal.

Answer 1

Answer: (A), (B)

The correct options are
A

The light ray will undergo total internal reflection.

B

The light ray will get reflected making an angle of ${60}^{\circ }$ with the normal.

Here ray of light is travelling from denser medium to rarer medium, there is a chance of total internal reflection.
Let the critical angle be $i_c$
Now using snell's law,
$\frac{sin{i}_c}{sinr}$ = $\frac{{\mu }_{air}}{{\mu }_{glass}}$

$\frac{sin{i}_c}{sin{90}^{\circ }}$ = $\frac{{\mu }_{air}}{{\mu }_{glass}}$

$i_c$ = ${\sin }^{-1}$ $\frac{{\mu }_{air}}{{\mu }_{glass}}$

$i_c$ = ${\sin }^{-1}$ $\frac{1}{{\mu }_{glass}}$

$i_c$ = ${\sin }^{-1}$ $\frac{1}{\sqrt{2}}$ ${\because \mu }_w$ = $\sqrt{2}$

We get $i_c={45}^{\circ }$

Here, angle of incidence $>$ $i_c$.

It is given that critical angle is ${45}^0$. Therefore if the angle of incidence is greater than ${45}^0$, then the light ray will undergo total internal reflection.

This reflected ray will follow the law of reflection i.e. the angle of incidence will be equal to the angle of reflection.

Therefore, the reflected ray will make ${60}^{\circ }$ angle with the normal.