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Reflection and Refraction

We mentioned in Section A that light traveling in a medium other than the vacuum has a speed slower than c. In fact it is given by

where n is the index of refraction. The chart at right gives some values of indices of refraction (do not memorize this chart, but note that n is always greater than 1).

substance n
vacuum 1
air   1
water   1.3
glass 1.5


When light traveling in one medium encounters a boundary to another medium, some of the light is reflected and some is transmitted into the second medium. If the incident light comes in at an angle, then the transmitted light is refracted, or bent, from its original direction. Figure 13-3 shows the wave fronts of light waves incident on glass from air.

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Figure 13-3

When working with diagrams of light waves, it is customary (and convenient) to use light rays rather than wave fronts. Rays point perpendicular to the front, and in the direction the light is going. Figure 13-4 is equivalent to Figure 13-3, using rays instead. Also, the normal to the surface (remember that "normal" means perpendicular) is shown as a dashed line.

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Figure 13-4

This is the most important figure of the chapter, so take a moment to study it. Notice the following:
  1. The angles are measured from the normal.
  2. The reflected angle θl is the same as the incident angle θi.
  3. The refracted angle θr (in glass) is smaller than the incident angle (in air). That is, the slower medium has the ray closer to the normal.

A light beam encounters a piece of glass as shown (Figure 13-5). Sketch the refracted path of the beam.


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Figure 13-5


In Figure 13-6 the normal is shown as a dashed line. The dotted line shows the path the beam would take if there were no glass. The ray bends, however, toward the normal. Now when it hits the other side of the glass, the beam bends away from the normal. The answer, then, is the solid line.


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Figure 13-6


Snell's law gives the refracted angle exactly:
Snell's Law
If a beam of light encounters a boundary, and if the beam is transmitted, the transmitted beam is refracted, or bent, according to
where /», and /», are the indices of refraction of the media housing the incident and refracted beams, and 0, and 0, are the angles of the beams, measured from the normal.
You will want to memorize it since it is in the MCAT study guide, but it is more important that you know how to use it.
A beam of light in air strikes the surface of pure hydrogen peroxide (n = 1.414) making an angle 30˚ with the normal to the surface.
  1. What angle does the reflected beam make with the normal?
  2. What angle does the transmitted beam make with the normal?
  1. The diagram for this problem is similar to Figure 13-4, with glass replaced by hydrogen peroxide. The reflected angle is the same as the incident angle, 30˚.
  2. The refracted angle is given by
nairsinθr = ndiasinθi
1.414 sinθr = 1.0 sin 30˚
sinθr = 0.354
θr = sin-1 0.354
= 20.7˚

where that last equation must be solved on a calculator.



A beam of light in a piece of diamond encounters an interface with air. The beam makes a 30˚ angle with the normal. What is the angle of refraction? (The index of refraction for diamond is 2.42.)



Figure 13-7 shows the diagram for this problem. Snell's law becomes


nairsinθr = ndiasinθi
1.0 sinθr = 2.42 sin 30
sinθr = 1.21


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Figure 13-7

Now we need to find an angle whose sine is 1.21. But wait a minute! There is no such thing as a sine which is greater than one. This equation has no solution. What is going on?

Figure 13-8 shows a beam of light in diamond with an angle of incidence of 20˚. The refracted ray bends away from the normal. Figure 13-9 shows a beam of light in diamond making a 24.4˚ angle with the normal. Once the beam gets into the air, it has bent so far from the normal that it is parallel with the surface. This angle, 24.4˚, is called the critical angle for a diamond-air interface. It is the incident angle for which the refracted angle is 90˚.


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Figure 13-8

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Figure 13-9


Figure 13-10 shows our situation with 30˚, but the refracted beam cannot bend any further away from the normal than it did for 24˚. In fact, there is no refracted ray in this case. This phenomenon is called total internal reflection, because all of the light stays in the diamond and none goes into the air.


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Figure 13-10


In a fiber optic cable, light travels down a light pipe with very little energy loss. You may have seen these cables in toys which were popular in the 1970s, where the tips of clear thin fibers light up with different colors. The cables have a high index of refraction, so that light gets totally internally reflected off the surface and thus does not leak out the sides. When it arrives at the tip, the light is transmitted into the air (Figure 13-11).


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Figure 13-11

Although we have discussed reflection and refraction only in the context of light waves, all waves in fact get reflected and refracted at boundaries. Figure 13-12 shows ocean waves coming toward the shore, as seen from above. As they approach, they encounter more and more shallow water, with smaller wave speed. Thus we would guess that they would bend toward the normal, so that the wave fronts would arrive roughly parallel to the shore. Compare Figure 13-12 to Figure 13-3. The difference here is that the air–glass boundary is sharp, whereas the boundary from deep to shallow water is gradual. Notice how the waves come in to the shore the next time you are at the beach.


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Figure 13-12

Also note (Figure 13-3 and Figure 13-12) that when a wave travels from one medium to another, the frequency stays the same, whereas the wavelength changes if the wave speed changes. This is easier to see with ocean waves than with waves of light. Imagine ocean waves in the deep portion of the ocean going up-down-up-down once every five seconds. If this goes on for a long time, it must be that the ocean waves arriving at shallow water go up-down-up-down once every five seconds as well. Think about it.

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