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9.2.4: Total Internal Reflection

A Limit to Snell's Law

Let us do another example of Snell’s law. We are going to look at light traveling from water (\(n_{water} = 1.33\)) to the air above (\(n_{air} = 1\)). If we are underwater we may choose \(\theta_{water}\) to be any angle between 0° and 90° by pointing our light source in the appropriate direction. Let us choose \(\theta_{water} = 60°\) as shown in the diagram below:

We now use Snell’s law to determine the angle \(\theta_{air}\) that the outgoing ray will emerge.

\[n_{water} \sin \theta_{water} = n_{air} \sin \theta_{air}\]

\[(1.33) \sin 60° = (1) \sin \theta_{air}\]

\[\implies \sin \theta_{air} = 1.152\]

This is obviously a problem, as we know that the sine of any real angle is between 1 and −1! Here Snell’s law does not give us an answer!

To get a slightly more intuitive feeling for the problem, recall that if we go from a slow medium to a fast medium the light “bends away from the normal”. As we increase the angle in the slow medium, the light refracting into the fast medium gets lower and lower to the boundary.  Eventually we "run out of room" and the light cannot bend into the fast medium.

A more direct way of understanding what is happening, if we go back to our wavefront model of refraction, is that it takes such a long time for a piece of a wavefront to make it out of the water and into the air, that any part of the wavefront that made it into the air would be forced to leave behind the wavefront in the water.

Total Internal Reflection and the Critical Angle

So we can understand why Snell’s law does not work anymore, but if the light cannot refract into the air, what happens to it? The answer is that all the light gets reflected instead. This is called total internal reflection, and it occurs whenever Snell’s law cannot be solved as in the problem above.  Notice that total internal reflection only occurs when light travels from a slow medium to a fast one. Light going from a fast medium to a slow one won't exhibit total internal reflection as there is always room to bend toward the normal.

The smallest angle for which total internal reflection occurs is called the critical angle \(\theta_c\). To find \(\theta_c\) we want to know when the refracted ray has bent away from the normal as far as it can go; namely \(\theta_{fast} = 90°\). For the case of water-to-air, we can find the critical angle as follows:

\[n_{water} \sin \theta_c = n_{air} \sin 90°\]

\[(1.33) \sin \theta_c = 1\]

\[\implies \theta_c = 48.8°\]

This means for \(\theta_{water} < 48.8°\) some of the incident light refracts into the air, but for \(\theta_{water} > 48.8°\) all the light is reflected back into the water. A picture from a swimming pool demonstrating total internal reflection and the critical angle is shown below.

This picture, from a camera underwater, shows the photographer's foot and the reflection of his foot off the surface of the pool.  Outside of the pool there's a pool house that can be seen above on the right.  The critical angle is the angle at which the pool house and things outside the pool can no longer be seen.

When the light rays are perpendicular to a surface almost all the light is transmitted or “refracted”. As we make the light rays hit at higher angles of incidence the amount of light refracted decreases, and the amount of light reflected increases. Once the angle of incidence is greater than or equal to the critical angle \(\theta_c\) none of the light is refracted; instead it is all reflected.  Above is a sequence of images that graphically demonstrates the increase of incident angle (the thickness of the line represents brightness of light).

Hands-On 

The simulation below offers a more engaging display of the topics discussed in this section (as well as in Refraction and Reflection). In "Intro" you can change the angle of a laser incident on a water surface.  Can you estimate what \(\theta_c\) is from your observations?  You can change the index of refraction for each media.  There's also an option to visualize the light as a wave instead of a ray.  Try playing around with these tools to better understand how \(\theta_c\) depends on the media used.