Total Internal Reflection and Critical Angle

A coin can be hidden in a glass of water

The coin cannot be seen from sideways but the same coin is visible from the top

This is because when there is water in the glass, light from the coin travels through the glass to our eyes at a particular angle

Therefore, beyond a particular angle, the coin becomes invisible to us

This is due to the behaviour of light, i.e the bending of light waves when it moves from one medium to another

The direction of bending depends upon the optical density of the medium

Now let us see the bending of light when it moves from a Denser medium to a Rarer medium

The above set up shows a source of light in a denser medium

If the incident ray is incident on the interface at a right angle, the ray does not undergo deviation

Now, the incident ray is incident at angle $\angle i_1$ on the interface

The ray suffers refraction, bends away from the normal by an angle $r_1$ and emerge as the refracted ray $B_1$

If we further increase the angle of incidence to $\angle i_2$, the incident ray suffers refraction

The refracted ray bends further away from the normal $N_2$ and lies along the plane

This refracted ray makes an angle $90^o$ with the normal

At this point, the angle of incidence is known as the Critical angle C

Now, if we increase the angle of incidence beyond the critical angle, the ray cannot refract anymore

Therefore, the ray comes back to the same medium after reflection

All the light from the glass is internally reflected, and this phenomenon is known as the Total Internal Reflection

Total Internal Reflection can be defined as the complete reflection of a light ray reaching an interface with a less dense medium when the angle of incidence exceeds the critical angle

For the total internal reflection to occur, the particular value for the angle of incidence could be calculated using Snell's Law

By Snell's law, ${}_w{\mu }_a=\frac{\sin i}{\sin r}$ Where,${}_w{\mu }_a$ is the Refractive index of Air with respect to Water

For $\angle i=\angle i_c$, $\angle r=90^o$ therefore we can write, $$\begin{gathered} {}_w{\mu }_a=\frac{\sin i_c}{1} \\ \therefore {}_w{\mu }_a=\sin i_c \end{gathered}$$

By the principle of reversibility, ${}_a{\mu }_w=\frac{1}{\sin i_c}$

The formula explains that for the given refractive index of the substance, what should be the critical angle

Revision

The angle of incidence beyond which rays of light passing through a denser medium to the surface of a rarer medium are totally reflected is known as the critical angle

The complete reflection of a light ray reaching an interface with a less dense medium when the angle of incidence exceeds the critical angle is known as Total Internal Reflection

The formula explains that for the given refractive index of the substance, what should be the critical angle. ${}_a{\mu }_w=\frac{1}{\sin i_c}$

The End