13.4: Refraction of Rays
By the end of this section, you will be able to:
- Determine the index of refraction, given the speed of light in a medium
- Describe how rays change direction upon entering a medium
- Apply the law of refraction in problem solving
You may often notice some odd things when looking into a fish tank. For example, you may see the same fish appearing to be in two different places (Figure \(\PageIndex{1}\)). This happens because light coming from the fish to you changes direction when it leaves the tank, and in this case, it can travel two different paths to get to your eyes. The changing of a light ray’s direction (loosely called bending) when it passes through substances of different refractive indices is called refraction and is related to changes in the speed of light, \(v=c/n\). Refraction is responsible for a tremendous range of optical phenomena, from the action of lenses to data transmission through optical fibers.
The Speed of Light in Matter
Today, the speed of light is known to great precision. In fact, the speed of light in a vacuum c is so important that it is accepted as one of the basic physical quantities and has the value
\[c=2.99792458\times 10^8m/s \equiv 3.00\times 10^8m/s \label{speed} \]
where the approximate value of 3.00×10 8 m/s is used whenever three-digit accuracy is sufficient.
The speed of light through matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, since its interaction varies with different atoms, crystal lattices, and other substructures. We can define a constant of a material that describes the speed of light in it, called the index of refraction n :
\[n=\dfrac{c}{v} \label{index} \]
where \(v\) is the observed speed of light in the material.
Since the speed of light is always less than c in matter and equals c only in a vacuum, the index of refraction is always greater than or equal to one; that is, n≥1. Table \(\PageIndex{1}\) gives the indices of refraction for some representative substances. The values are listed for a particular wavelength of light, because they vary slightly with wavelength. (This can have important effects, such as colors separated by a prism, as we will see in Dispersion .) Note that for gases, n is close to 1.0. This seems reasonable, since atoms in gases are widely separated, and light travels at c in the vacuum between atoms. It is common to take \(n=1\) for gases unless great precision is needed. Although the speed of light v in a medium varies considerably from its value c in a vacuum, it is still a large speed.
Calculate the speed of light in zircon, a material used in jewelry to imitate diamond.
Strategy
We can calculate the speed of light in a material \(v\) from the index of refraction \(n\) of the material, using Equation \red{index}
Solution
Rearranging Equation \ref{index} for \(v\) gives us
\[v=\dfrac{c}{n}. \nonumber \]
The index of refraction for zircon is given as 1.923 in Table \(\PageIndex{1}\), and \(c\) is given in Equation \ref{speed}. Entering these values in the equation gives
\[\begin{align*} v &=\dfrac{3.00×10^8m/s}{1.923} \\[4pt] &=1.56×10^8m/s. \end{align*} \nonumber \]
Significance
This speed is slightly larger than half the speed of light in a vacuum and is still high compared with speeds we normally experience. The only substance listed in Table \(\PageIndex{1}\) that has a greater index of refraction than zircon is diamond. We shall see later that the large index of refraction for zircon makes it sparkle more than glass, but less than diamond.
Table \(\PageIndex{1}\) shows that ethanol and fresh water have very similar indices of refraction. By what percentage do the speeds of light in these liquids differ?
- Answer
-
2.1% (to two significant figures)
Figure \(\PageIndex{2}\) shows how a ray of light changes direction when it passes from one medium to another. As before, the angles are measured relative to a perpendicular to the surface at the point where the light ray crosses it. (Some of the incident light is reflected from the surface, but for now we concentrate on the light that is transmitted.) The change in direction of the light ray depends on the relative values of the indices of refraction of the two media involved. In the situations shown, medium 2 has a greater index of refraction than medium 1. Note that as shown in Figure \(\PageIndex{1a}\), the direction of the ray moves closer to the perpendicular when it progresses from a medium with a lower index of refraction to one with a higher index of refraction. Conversely, as shown in Figure \(\PageIndex{1b}\), the direction of the ray moves away from the perpendicular when it progresses from a medium with a higher index of refraction to one with a lower index of refraction. The path is exactly reversible.
The amount that a light ray changes its direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction and thus a large change in angle. The exact mathematical relationship is the law of refraction, or Snell’s law, after the Dutch mathematician Willebrord Snell (1591–1626), who discovered it in 1621. The law of refraction is stated in equation form as
\[n_1 \, \sin \, θ_1=n_2 \, \sin \, θ_2. \label{snell's law} \]
Here (n_1\) and \(n_2\) are the indices of refraction for media 1 and 2, and \(θ_1\) and \(θ_2\) are the angles between the rays and the perpendicular in media 1 and 2. The incoming ray is called the incident ray, the outgoing ray is called the refracted ray, and the associated angles are the incident angle and the refracted angle, respectively.
Snell’s experiments showed that the law of refraction is obeyed and that a characteristic index of refraction \(n\) could be assigned to a given medium and its value measured. Snell was not aware that the speed of light varied in different media, a key fact used when we derive the law of refraction theoretically using Huygens’s Principle .
Find the index of refraction for medium 2 in Figure \(\PageIndex{1a}\), assuming medium 1 is air and given that the incident angle is 30.0° and the angle of refraction is 22.0°.
Strategy
The index of refraction for air is taken to be 1 in most cases (and up to four significant figures, it is 1.000). Thus, \(n_1=1.00\) here. From the given information, \(θ_1=30.0°\) and \(θ_2=22.0°\). With this information, the only unknown in Snell’s law is \(n_2\), so we can use Snell’s law (Equation \ref{snell's law}) to find it.
Solution
From Snell’s law (Equation \ref{snell's law}), we have
\[\begin{align*} n_1\sin θ_1 &=n_2 \sin θ_2 \\[4pt] n_2 &= n_1\dfrac{\sin θ_1}{\sin θ_2}. \end{align*} \nonumber \]
Entering known values,
\[\begin{align*} n_2 &=1.00 \dfrac{\sin 30.0°}{\sin 22.0°} \\[4pt] &= \dfrac{0.500}{0.375} \\[4pt] &=1.33. \end{align*} \nonumber \]
Significance
This is the index of refraction for water, and Snell could have determined it by measuring the angles and performing this calculation. He would then have found 1.33 to be the appropriate index of refraction for water in all other situations, such as when a ray passes from water to glass. Today, we can verify that the index of refraction is related to the speed of light in a medium by measuring that speed directly.
Explore bending of light between two media with different indices of refraction. Use the “Intro” simulation and see how changing from air to water to glass changes the bending angle. Use the protractor tool to measure the angles and see if you can recreate the configuration in Example \(\PageIndex{1}\). Also by measurement, confirm that the angle of reflection equals the angle of incidence.
Suppose that in a situation like that in Example \(\PageIndex{1}\), light goes from air to diamond and that the incident angle is 30.0°. Calculate the angle of refraction θ 2 in the diamond.
Strategy
Again, the index of refraction for air is taken to be n 1 =1.00, and we are given θ 1 =30.0°. We can look up the index of refraction for diamond , finding n 2 =2.419. The only unknown in Snell’s law is \(θ_2\), which we wish to determine.
Solution
Solving Snell’s law (Equation \ref{snell's law}) for \(\sin θ_2\) yields
\[\sin θ_2=\frac{n_1}{n_2}\sin θ_1. \nonumber \]
Entering known values,
\[\sin θ_2=\frac{1.00}{2.419}\sin30.0°=(0.413)(0.500)=0.207. \nonumber \]
The angle is thus
\[θ_2=\sin^{−1}(0.207)=11.9°. \nonumber \]
Significance
For the same 30.0° angle of incidence, the angle of refraction in diamond is significantly smaller than in water (11.9° rather than 22.0°—see Example \(\PageIndex{2}\)). This means there is a larger change in direction in diamond. The cause of a large change in direction is a large change in the index of refraction (or speed). In general, the larger the change in speed, the greater the effect on the direction of the ray.
The solid with the next highest index of refraction after diamond is zircon. If the diamond in Example \(\PageIndex{2}\) were replaced with a piece of zircon, what would be the new angle of refraction?
- Answer
-
15.1°