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9.2.3: Refraction

Medium Affects Wavelength

We have already discussed that the speed of a wave is determined by the medium, and this is true for the speed of light as well. Recall that the frequency of a wave is set by the source, so the frequency of a wave does not change as it travels into a new medium.  This is true also for light waves traveling through different media

Because we know the wave speed and frequency are related by \(v_{wave} = f \lambda\), we see that if \(v_{wave}\) changes \(\lambda\) must also change.  In other words, the wavelength of light changes when it travels into a different medium with a different allowed speed of light.

Another way of understanding this is as follows: if \(v_{wave}\) is small, then the wave cannot travel very far in one period so \(\lambda\) is small, and if \(v_{wave}\) is large then one peak can travel much further in one period so \(\lambda\) is large.

Normal Incidence

Let us consider the concrete case of light traveling from air into water.  First, we establish that light travels faster in air than it does in water. Imagine the plane wavefronts of the light traveling from air into water in the same direction as the normal, so the wavefronts are parallel with the surface. This is called normal incidence, as the light rays are travelling along the normal of the air-water boundary. We note that light travels faster in air than it does in water, and this makes the wavelength of light in water shorter.

How Refraction Occurs

The situation becomes much more interesting if the wavefronts of the light rays are not lined up exactly with the air-water boundary as shown below.

Imagine what happens to one wavefront as it enters the water.  Part of a wavefront enters the water and slows down while the rest of the wavefront stays in the air at its original speed. The wavefront in the air tries to speed ahead the wavefront in the water, but they still have to join smoothly at the boundary. This causes the whole wavefront to bend as demonstrated below.

Traveling plane wave entering a region of lower wave velocity at an angle, illustrating the decrease in wavelength and change of direction (refraction) that results. Image used with permission (CC-BY-SA; Richard F. Lyon)

A useful analogy is to consider a car traveling from the road (a fast medium) to mud (a slow medium).  As the car travels, one tire goes forward faster than the other, which causes the entire car to turn.   This is demonstrated visually below.  Note that this analogy works for both the cases where the car goes from a fast medium to a slow one or from a slow medium to a fast one.

Going back to our original picture, we can draw in rays perpendicular to our wavefronts, and notice that the direction of the light bends as we go from air to water. This bending occurs for any two different media in which the light waves have different speeds. This bending of light as it goes from one medium into another is called refraction. Notice that, although the light clearly bends, light rays travel in straight lines within each medium.

Combining Reflection and Refraction

You may wonder when a light ray hits a surface, how can we tell if it is going to be reflected or refracted? The answer is that a light ray is typically both reflected and refracted (we discuss this more in Total Internal Reflection).

You might already have some familiarity with this from your experience with swimming pools. It is possible to see the sun from inside a swimming pool, so we know that light from the sun must be able to make it into the water. Therefore the sun’s rays are refracting as they enter the water. Someone standing on the side of the pool can also see the reflection of the sun on the water’s surface (known as the “glare”), so the sun’s rays must also be reflecting off the surface of the pool. There is no contradiction here – when the sun’s rays hit the surface some rays reflect and other rays refract.

Because energy must be conserved, reflected and refracted rays have less energy than the incident ray.  For the moment we will simply concentrate on the refracted ray and omit the reflected ray from our discussion, but it is there nevertheless.

In refraction it is common to talk about the “fast” medium (with the high wave speed) and the “slow” medium (with the lower wave speed). In the case of light going from air to water, the fast medium is air and the slow medium is water. The above examples of refraction showed that when light travels from a fast medium to a slow medium the light rays bend toward the normal.

Exercise

Show that when light travels from a slow medium to a fast medium, the light rays bend away from the normal.  The diagram below helps to illustrate what is meant when we say the rays bend "toward" or "away" form the normal.

Refractive Indices

That is all the qualitative information we need about refraction. We now turn to the quantitative task of determining precisely which way a refracted ray travels as it goes from one medium to another. We know that the amount of bending depends on the speed of the wave in the medium. For convenience we introduce a new concept, the refractive index.  The refractive index \(n\) for light in a particular medium is defined such that \[n\equiv \dfrac{\text{speed of light in vacuum}}{\text{speed of light in medium}} = \dfrac{c}{v_{wave}}\]

The reason for this is that the speed of light in materials is typically \(10^7\text{ - }10^8\text{ m/s}\), while the \(n\) for most materials is between one and five. The utility of the refractive index is that the values of \(n\) are easier to use than the values for \(v_{wave}\). From the definition of the refractive index, we know three things:

  1. \(n_{medium} \geq 1\), because nothing can travel faster than the speed of light in a vacuum.
  2. Given two media, the slower medium will have the larger \(n\) value and the faster one will have the smaller \(n\) value.
  3. \(n_{vacuum} = \dfrac{c}{c} = 1\).

The refractive indices for some materials are given in this table

Material \(n = c/v_{medium}\) Material \(n = c/v_{medium}\)
Vacuum 1 (exact) Silicon 3.5
Air 1.0003 Germanium 4.0
Water 1.33 Diamond 2.42
Glass (crown) 1.50 - 1.62 Eye 1.3
Glass (flint) 1.57 - 1.75 Eye lens 1.41

Snell's Law

With the definition of refractive index we can now give a quantitative description of refraction. We will call the refractive index in one of the medium \(n_1\) and the angle of the light ray in that medium is \(\theta_1\), and for the second medium we will use \(n_2\) and \(\theta_2\). All these quantities are related by Snell’s law:

\[n_1 \sin \theta_1 = n_2 \sin \theta_2\]

This result is simply presented above, but it can actually be derived from what we already know about waves (a derivation is presented in the summary).

For your work, it will probably be more convenient to ignore the labels “1” and “2” and instead use the names of the media. An example of applying Snell's Law to light traveling from water to air is presented below:

Example 3

If we placed a point source of light in a calm pool, how would the light bend coming into the air?

Solution

We know that rays come off the light source in all different directions, but we will only consider the light that exits the surface of the pool.

We know that the light ray that is at normal incidence (\(\theta_{water} = 0\)) will pass through without bending.  This result, which we described earlier, follows clearly from Snell's Law.

By applying Snell’s law separately to rays pointing in different directions, we see that as we transition from normal incidence to high \(\theta_{inc}\) values, the bending becomes more severe. We illustrate multiple refracted rays in the picture below.

Example 4

A ray of light (in air) hits a glass prism horizontally as shown below.  The glass has a refractive index \(n= 1.5\). At what angle does the light refract inside the glass?

Solution

We know from a previous example (and trigonometry) that the incoming light ray is at an angle \(\theta_{air} = 60°\) from the normal. The refractive index for air is about one, so we can use Snell’s law:

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

\[\implies \sin \theta_{glass} = \dfrac{n_{air}}{n_{glass}} \sin \theta_{air} = \dfrac{1}{1.5} \times \sin 60° = 0.58\]

We can use the inverse sine function (\(\sin^{-1}\)) to find \(\theta_{glass} = 35°\).  The path of the ray looks like this

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