Question

A ray of light travels from a medium to air at an angle of incidence ${{45}^{\circ }}$. The critical angle of medium w.r.t air is ${{40}^{\circ }}$. Find the deviation suffered by the ray.

Answer 1

Hint: Critical angle is the minimum angle required for reflection of light when the light passes from a denser medium and rarer medium. If the angle of incidence is greater than the critical angle, the light ray reflects into the same medium instead of refraction.

Formula used:
$\sin {{\theta }_{c}}=\dfrac{1}{\mu }$
${{\mu }_{i}}\sin i={{\mu }_{r}}\sin r$

Complete step by step answer:
Let us first understand what is meant by critical angle.
We know that when a ray of light passes from one medium into another medium, the light refracts at the interface at the two mediums.
However, when a ray of light passes from a denser medium into a rarer medium, the light ray may refract or may be reflected into the same medium of incidence. The reflection of light at the interface is called total internal reflection.
Whether the light ray will refract or reflect depends on the angle of incidence. The minimum angle required for total internal reflection is called critical angle.
It is given that the critical angle for the medium and air is ${{40}^{\circ }}$.
It is given that the angle of incidence is ${{45}^{\circ }}$. Therefore, the ray will suffer total internal reflection as shown.

We know that when light reflects, the angles of incidence and refraction are equal. This means that the angle of reflection is ${{45}^{\circ }}$.
From the given figure we get that the angle deviates by an angle of ${{90}^{\circ }}$.

Note:
Let us assume that the light ray refracts instead of reflection. Then we have to apply Snell’s law.
The critical angle when the light passes from a denser medium of refractive index $\mu $ ($\mu $>1) into air is given as ${{\theta }_{c}}={{\sin }^{-1}}\left( \dfrac{1}{\mu } \right)$.
$\Rightarrow \sin {{\theta }_{c}}=\dfrac{1}{\mu }$.
It is given that ${{\theta }_{c}}={{40}^{\circ }}$.
$\Rightarrow \sin \left( {{40}^{\circ }} \right)=\dfrac{1}{\mu }$
$\Rightarrow \mu =\dfrac{1}{\sin \left( {{40}^{\circ }} \right)}$
From Snell’s law we get that ${{\mu }_{i}}\sin i={{\mu }_{r}}\sin r$ …. (i),
where ${{\mu }_{i}}$ is the refractive index of the medium in which the light is incident, ${{\mu }_{r}}$ is the refractive index of the medium in which the light is refracted.
In this case, ${{\mu }_{i}}=\mu =\dfrac{1}{\sin {{40}^{\circ }}}$, ${{\mu }_{r}}=1$ and $i={{45}^{\circ }}$.
Substitute the values of ${{\mu }_{i}}$, ${{\mu }_{r}}$ and i in equation (i).
$\Rightarrow \left( \dfrac{1}{\sin {{40}^{\circ }}} \right)\sin {{45}^{\circ }}=1.\sin r$
$\Rightarrow \sin r=\dfrac{\sin {{45}^{\circ }}}{\sin {{40}^{\circ }}}=1.1$
However, sine of an angle is always between -1 and 1. Hence, sin(r) cannot be greater than 1. This proves that the light ray does suffer refraction.