Critical Angles and Total Internal Reflection
Total internal reflection
Optional Video on Total internal reflection
You may have noticed when experimenting with ray boxes and glass blocks in the previous section that sometimes, when you changed the angle of incidence of the light, it was not refracted out into the air, but was reflected back through the block. When the entire incident light ray travelling through an optically denser medium is reflected back at the boundary between that medium and another of lower optical density, instead of passing through and being refracted, this is called total internal reflection.
As we increase the angle of incidence, we reach a point where the angle of refraction is \(\text{90}\)\(\text{°}\) and the refracted ray travels along the boundary of the two media. This angle of incidence is called the critical angle.
Definition: Critical angle
The critical angle is the angle of incidence where the angle of refraction is \(\text{90}\)\(\text{°}\). The light must travel from an optically more dense medium to an optically less dense medium.
When the angle of incidence is equal to the critical angle, the angle of refraction is equal to \(\text{90}\)\(\text{°}\).
If the angle of incidence is bigger than this critical angle, the refracted ray will not emerge from the medium, but will be reflected back into the medium. This is called total internal reflection.
The conditions for total internal reflection are:
- light is travelling from an optically denser medium (higher refractive index) to an optically less dense medium (lower refractive index).
- the angle of incidence is greater than the critical angle.
When the angle of incidence is greater than the critical angle, the light ray is reflected at the boundary of the two media and total internal reflection occurs.
Each pair of media have their own unique critical angle. For example, the critical angle for light moving from glass to air is \(\text{42}\)\(\text{°}\), and that of water to air is \(\text{48.8}\)\(\text{°}\).
A recommended experiment for informal assessment is also included. This covers determining the critical angle for light travelling through a rectangular glass block. Learners will need a rectangular glass block, a \(\text{360}\)\(\text{°}\) protractor, pencil, paper, ruler and a ray box. Learners should all get similar results at the end of the experiment.
Optional Experiment: The critical angle of a rectangular glass block
Aim
To determine the critical angle for a rectangular glass block.
Apparatus
rectangular glass block, ray box, \(\text{360}\)\(\text{°}\) protractor, paper, pencil, ruler
Method
-
Place the glass block in the middle of the piece of paper and draw around the outside of the block with your pencil to make its outline.
-
Turn on the ray box and aim the light ray into the left side of the glass block. Adjust the angle at which the light strikes the glass block until you see the refracted light ray travelling along the top edge of the glass block (i.e. the angle of refraction is \(\text{90}\)\(\text{°}\)). This situation should look something like the following diagram:
-
Draw a dot on the paper at the point where the light enters the glass block from the ray box. Then draw a dot on the paper at the point where the light is refracted (at the top of the glass block).
-
Turn off the ray box and remove the glass block from the paper.
-
Now use your ruler to draw a line between the two dots. This line represents the incident light ray.
-
When the angle of refraction is \(\text{90}\)\(\text{°}\), the angle of incidence is equal to the critical angle. Therefore to determine the critical angle, we need to measure this angle of incidence. Do this using your protractor.
-
Compare your answer with the values other members of your class obtain and discuss why they might not be identical (although they should be similar!).
Calculating the critical angle
Instead of always having to measure the critical angles of different materials, it is possible to calculate the critical angle at the surface between two media using Snell's Law. To recap, Snell's Law states: \[n_1 \sin \theta_1 = n_2 \sin \theta_2\] where \(n_1\) is the refractive index of material \(\text{1}\), \(n_2\) is the refractive index of material \(\text{2}\), \(\theta_1\) is the angle of incidence and \(\theta_2\) is the angle of refraction. For total internal reflection we know that the angle of incidence is the critical angle. So, \[\theta_1 = \theta_c.\]
However, we also know that the angle of refraction at the critical angle is \(\text{90}\)\(\text{°}\). So we have: \[\theta_2=90^{\circ}.\]
We can then write Snell's Law as: \[n_1 \sin{\theta_c} = n_2 \sin{90^{\circ}}\]
Solving for \(\theta_c\) gives: \begin{align*} n_1 \sin \theta_c &= n_2 \sin \text{90}\text{°} \\ \sin \theta_c &= \frac{n_2}{n_1}(1) \end{align*} \[\boxed{\therefore \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)}\]
Tip:
Remember that for total internal reflection the incident ray is always in the denser medium!
Example: Critical Angle
Question
Given that the refractive indices of air and water are \(\text{1.00}\) and \(\text{1.33}\) respectively, find the critical angle.
Step 1: Determine how to approach the problem
We can use Snell's law to determine the critical angle since we know that when the angle of incidence equals the critical angle, the angle of refraction is \(\text{90}\)\(\text{°}\).
Step 2: Solve the problem
\begin{align*} n_1 \sin{\theta_c} &= n_2 \sin{90^{\circ}}\\ \theta_c &=\sin^{-1}\left(\frac{n_2}{n_1}\right) \\ &=\sin^{-1}\left(\frac{1}{\text{1.33}}\right) \\ &= \text{48.8}\text{°} \end{align*}
Step 3: Write the final answer
The critical angle for light travelling from water to air is \(\text{48.8}\)\(\text{°}\).
Example:
Question
Complete the following ray diagrams to show the path of light in each situation.
Step 1: Identify what is given and what is asked
The critical angle for water is \(\text{48.8}\)\(\text{°}\)
We are asked to complete the diagrams.
For incident angles smaller than \(\text{48.8}\)\(\text{°}\) refraction will occur.
For incident angles greater than \(\text{48.8}\)\(\text{°}\) total internal reflection will occur.
For incident angles equal to \(\text{48.8}\)\(\text{°}\) refraction will occur at \(\text{90}\)\(\text{°}\).
The light must travel from a medium with a higher refractive index (higher optical density) to a medium with lower refractive index (lower optical density).
Step 2: Complete the diagrams
Refraction occurs (ray is bent away from the normal).
Total internal reflection occurs.
\(\theta_c = \text{48.8}\text{°}.\)
Refraction towards the normal (air is less dense than water).
This lesson is part of:
Optics and Optical Phenomena