Total Internal Reflection

Does refraction work for every possible angle of incidence? When light moves from a more optically dense material to a less optically dense material, for angles greater than a certain value, there will be no refracted ray but rather the light will reflect within the first medium.

Total internal reflection is the phenomenon in which light bounces off of the boundary between a more optically dense material and a less optically dense material. The light reflects within the more optically dense material. The angle of incidence at which the refracted ray travels along the boundary separating the two media is called the critical angle. If the angle of incidence is greater than the critical angle then total internal reflection occurs.

Credits: Wikipedia

The critical angle for light moving from glass to air can be found if we use the refractive index of the air relative to glass: $$ \begin{equation}\begin{aligned} _gn_a=\frac{\sin\theta_g}{\sin\theta_a}\\ \end{aligned}\end{equation} $$

Notice the order of the $a$’s and $g$’s in the formula

$c$ is the angle of incidence when $90\degree$ is the angle of refraction (since the light travels along the boundary which is $90\degree$ measured from the normal). $$ \begin{equation}\begin{aligned} _gn_a&=\frac{\sin c}{\sin90\degree}\\ &=\frac{\sin c}{1}\\ &=\sin c\\ \therefore c&=\sin^{-1}(_gn_a)\\ \end{aligned}\end{equation} $$ This is good to know but it is common to have the refractive index of glass relative to air (as air has an absolute refractive index of $1$). The refractive index of air relative to glass ($_gn_a$) is the reciprocal of that of glass relative to air ($_an_g$):

$$ \begin{equation}\begin{aligned} c=\sin^{-1}(\frac{1}{_an_g})\\ \end{aligned}\end{equation} $$

The refractive index of glass relative to air (or simply the refractive index of glass) is about $1.5$ on average thus the critical angle is expected to be:

$$ \begin{equation}\begin{aligned} c&=\sin^{-1}(\frac{1}{1.5})\\ &=41.8\degree\\ \end{aligned}\end{equation} $$

Mock lab

1. Follow the link here
2. Select ‘Intro’
3. Align the protractor vertically with the broken line (the normal) and horizontally with the boundary between the air and the water
4. Turn on the laser
5. Try different angles of incidence and measure and record the respective angles of refraction
6. Set the material above the boundary to ‘Glass’ and the material below the boundary to ‘Air’
7. Repeat step $5$ for this pair of media
8. Try angles of incidence lesser than and greater than $41.8\degree$
9. Use the critical angle formula to figure out the critical angle for light exiting water into air and verify this value using the simulation

A more structured approach

1. After finding the angles of incidence and their respective angles of refraction, we can calculate the sines of these:

2. Plot a graph of sine of the angle of incidence versus the sine of the angle of refraction:

$$ \begin{equation}\begin{aligned} n&=\frac{\sin\theta_i}{\sin\theta_r}\\ \color{royalblue}\\ \sin\theta_i&=\color{purple}n\color{black}\times \color{orange}\sin\theta_r\color{gray}+0\\ \color{royalblue}y&=\color{purple}m\color{orange}x\color{grey}+c\\ \end{aligned}\end{equation} $$

As we see here, the gradient of the graph will be the refractive index of the second medium relative to the first medium and thus the final step is: 3. Calculate the gradient of the graph

Mission details

• How do refraction and total internal reflection affect the way in which fish are able to see while in water?

Video on archerfish:

• What are some applications of total internal reflection?