The medium whose refractive index is greater is known as optically denser medium whereas one which has smaller value is known as optically rarer medium.

In the figure shown light travels from medium I to medium II, n_{1} and n_{2} corresponds to their refractive indices and n_{1} > n_{2}

For any angle of incidence θ1 in medium I corresponding angle of refraction in medium II is θ_{2}

n_{1} sin θ_{1} = n_{2} sin θ_{2}

$ \displaystyle sin\theta_2 = \frac{n_1}{n_2}sin\theta_1 $

as n_{1} > n_{2}, sin θ_{2} > sin θ_{1} and hence θ_{2} > θ_{1}

In this way for a certain value of θ_{1 }, θ_{2} will be equal to 90°.

This value of θ_{1} is known as critical angle(θ_{C}).

If θ_{1} > critical angle, θ_{2} should be greater than 90°.

This implies that the ray will return to the previous medium.

For the angle of incidence greater than critical angle light does not transmit into medium II rather it reflects back to the medium I .

This phenomenon is called total internal reflection.

If θc is the critical angle then

$ \displaystyle sin90 = \frac{n_1}{n_2}sin\theta_c $

$ \displaystyle sin\theta_c = \frac{n_2}{n_1} $

If medium II is air then n_{2} ≃ 1.

$ \displaystyle sin\theta_c = \frac{1}{n_1} $

Example : A small coin is placed at the bottom of a cylindrical vessel of radius R and depth h. If a transparent liquid of refractive index n completely filled into the cylinder , find the minimum fraction of the area that should be covered in order not to see the coin , if viewed form top of vessel.

Solution : When the rays coming from the coin incident at an angle θ ≥ θ_{c}, they will be totally reflected.

=> Minimum area that should be covered = A = πr^{2}

where r = radius of the circular aperture.

and , r = h tan θ_{c}

when θ_{c} can be given by the formula

$ \displaystyle sin\theta_c = \frac{n_a}{n} \frac{1}{n}$

$ \displaystyle tan\theta_c = \frac{1}{\sqrt{n^2 – 1}} $

$ \displaystyle r = \frac{h}{\sqrt{n^2 – 1}} $

$ \displaystyle A = \pi r^2 = \frac{\pi h^2}{n^2 – 1} $

The total area of the aperture through which coin can be seen

A’ = πR^{2}

The fraction of aperture through which coin can covered

$ \displaystyle \eta = \frac{A}{A’} = \frac{\pi h^2}{n^2 – 1}/ \pi R^2 $

$ \displaystyle \eta = \frac{h^2}{(n^2 – 1)R^2} $

Exercise : A black spot is situated at the body center of a transparent glass (n = 1.5) cube of edge 0.1 m, placed on a table. What minimum fraction of the cube must be covered in order not to visualise the spot.