Video: Comparing Reflection and Refraction

Light strikes the boundary between air and glass at an angle. Some of the light is reflected and some of it is refracted. Which of the following properties of the light is the same for both the incident ray and the refracted ray? [A] Power transmitted [B] Direction [C] Frequency [D] Speed [E] Wavelength

09:47

Video Transcript

Light strikes the boundary between air and glass at an angle. Some of the light is reflected and some of it is refracted. Which of the following properties of the light is the same for both the incident ray and the refracted ray? (a) Power transmitted, (b) direction, (c) frequency, (d) speed, (e) wavelength.

Let’s draw a picture of the situation described in the question. Here’s the boundary. And there is air on one side and glass on the other. A ray of light is incident on the boundary, and it is incident at an angle. In the diagram, we’ve labeled this angle with the symbol πœƒ 𝑖 measured from a line perpendicular to the boundary. Some of this light is then reflected, that is, bounces back into the air. And some of the light is refracted, that is, transmitted into the glass possibly with a change of direction. The question wants to know which of the properties given is the same for both the incident ray and the refracted ray. So let’s take a look at these properties.

Choice (a) is power transmitted. Because beams of light carry energy, they also transfer power, which is just a transfer of energy with respect to time. Therefore, since we’re interested in power before the light hits the boundary and after the light hits the boundary and power is directly related to energy. It makes sense to apply the conservation of energy, which states that the total initial energy is equal to the total final energy. Let’s call the power transmitted by the incident ray 𝑝 zero, by the reflected ray 𝑝 one, and by the refracted ray 𝑝 two. Because there’s no absorption at the boundary and energy is conserved, the energy carried per unit time to the boundary by the incident ray must exactly equal the total energy carried away from the boundary per unit time by the reflected and refracted rays.

Since carrying energy to or away from the boundary is the same as transmitting power, we can write for our conservation of energy that 𝑝 zero, the power transmitted by the incident ray, is equal to 𝑝 one plus 𝑝 two, the total power transmitted by the reflected and refracted rays. If we subtract 𝑝 one from both sides, we get that 𝑝 two, the power of the refracted ray, is equal to 𝑝 zero minus 𝑝 one, the power of the incident ray minus the power of the reflected ray. Now we make a key observation. From the question itself, we know that the reflected ray must exist, which means that it must have energy and must be transmitting power. So 𝑝 one is greater than zero. However, if 𝑝 one is greater than zero, it follows that 𝑝 zero is not equal to 𝑝 zero minus 𝑝 one because 𝑝 one is greater than zero. So 𝑝 zero minus 𝑝 one must be less than 𝑝 zero.

Note that we have a different expression for 𝑝 zero minus 𝑝 one; namely, 𝑝 zero minus 𝑝 one is equal to 𝑝 two. So let’s substitute this in to 𝑝 zero is not equal to 𝑝 zero minus 𝑝 one. Doing the substitution, we find 𝑝 zero is not equal to 𝑝 two. And this is exactly the sort of relationship we’re looking for. The power transmitted by the incident ray is not the same as the power transmitted by the refracted ray. And (a) is not the right answer.

Let’s move on to choice (b) direction. The direction of the incident and refracted rays can be specified by their angle relative to the line perpendicular to the boundary. We already had πœƒ 𝑖 for the incident ray on our diagram from before. And now we’ve added πœƒ 𝑓 for the angle between the refracted ray and the line perpendicular to the boundary. To relate these two angles and find out if they’re the same or not, we’ll use Snell’s law. Snell’s law relates the angle of an incident ray to the angle of a refracted ray through a quantity known as the index of refraction. The index of refraction is a number that relates the speed of light in a particular medium to the speed of light in vacuum. 𝑛 π‘Ž, the index of refraction of air, is approximately one. And 𝑛 𝑔, the index of refraction of glass, is approximately 1.3.

The exact numerical values are not terribly important. But what is important is that the index of refraction of air is different from the index of refraction of glass. Letting the subscript one denote that we’re talking about the incident ray and the subscript two denote that we’re talking about the refracted ray, let’s plug in the variables from our diagram into Snell’s law. We get 𝑛 π‘Ž sin πœƒ 𝑖, the index of refraction in air times the sin of the angle of the incident ray, is equal to 𝑛 𝑔 sin πœƒ 𝑓, the index of refraction in glass times the sin of the angle of the refracted ray.

Let’s now use the fact that 𝑛 π‘Ž is not equal to 𝑛 𝑔. If 𝑛 π‘Ž is not equal to 𝑛 𝑔, it follows that sin πœƒ 𝑖 cannot be equal to sin πœƒ 𝑓 because, otherwise, Snell’s law would lead to a contradiction. But if sin πœƒ 𝑖 is not equal to sin πœƒ 𝑓, then certainly πœƒ 𝑖 is not equal to πœƒ 𝑓. But if πœƒ 𝑖 is not equal to πœƒ 𝑓, the incident and refracted rays have different directions, and (b) is not the right answer. We can use the fact that the index of refraction of air is not equal to the index of refraction of glass to eliminate another one of the answer choices by recalling how exactly the index of refraction relates the speed of light in a material to the speed of light in vacuum.

The index of refraction can be defined as the speed of light in vacuum, which we’ve represented with the symbol 𝑐 without any subscripts, divided by the speed of light in the particular material of interest. Thus, for example, the index of refraction of air is about one because the speed of light in air is about the same as the speed of light in vacuum, whereas the speed of light in glass is about 75 percent of that in vacuum. So the index of refraction of glass is about 1.3. Let’s now plug this into our inequality 𝑛 π‘Ž is not equal to 𝑛 𝑔. Letting 𝑐 π‘Ž be the speed of light in air and 𝑐 𝑔 be the speed of light in glass, we have that speed of light in vacuum divided by speed of light in air is not equal to speed of light in vacuum divided by speed of light in glass. Both the left- and right-hand sides have a factor of speed of light in vacuum. So let’s multiply both sides by one over 𝑐.

On both sides, 𝑐 divided by 𝑐 is one. And so we’re left with one divided by the speed of light in air is not equal to one divided by the speed of light in glass. But if the reciprocal of two quantities are not equal, then the quantities themselves are also not equal. We therefore find that the speed of light in air is not equal to the speed of light in glass. And so the incident and refracted rays have different speeds. And choice (d) is not correct. We’re left with choices (c) frequency and (e) wavelength. Let’s take a closer look at the boundary where the incident and refracted rays meet and see if we can come up with the condition for the frequency to be either the same or different.

Here, we have redrawn the boundary, the incident ray and the refracted ray, ignoring the reflected ray and the other details from our previous drawing. Light is an electromagnetic wave. So we’ve drawn these two rays as squiggles instead of straight lines to emphasize that the electric field of each ray is oscillating. The squiggles represent the amplitude of the electric field. And in our picture, an oscillation would look like the squiggles that are towards the bottom left moving towards the top right or the squiggles that are towards the top right move towards the bottom left.

Let’s now examine what happens to these oscillations at the boundary where the two rays meet. The other fact that we’ll need to know to analyze this situation is that the electric field is continuous across a boundary. That is, it’s the same on both sides of the boundary. Since the electric field is continuous, the incident and refracted rays must oscillate in tandem. So when the incident ray is oscillating towards the top right, the refracted ray must as well. And the same is true if the oscillation is towards the bottom left.

However, because of the continuity of the electric field, for all times, the incident and refracted rays need to oscillate in the same direction as each other, which means they must oscillate at the same rate. If the two fields oscillated at different rates, at some time, we would find the electric field of the incident ray at the boundary oscillating towards the upper right, while the electric field of the refracted ray at the boundary oscillated towards lower left. But this would literally result in the electric field being pulled apart at the boundary.

The rate of oscillation is just a frequency. So we see that both rays must have the same frequency at the boundary. This, in turn, means that both rays must have the same frequency everywhere because the frequency is constant within a medium. So the frequency of the incident ray is the same as the frequency of the incident ray at the boundary. And the frequency of the refracted ray is the same as the frequency of the refracted ray at the boundary. But those two frequencies at the boundary are equal. So because the electric field is continuous at the boundary, we find that the correct answer is choice (c) the frequency.

As a final note, let’s confirm that wavelength would indeed be different. We’ll do this using the equation speed of light is equal to frequency times wavelength, where wavelength is represented with the symbol πœ†. Let’s substitute this equation into our inequality speed of light in air is not equal to speed of light in glass. If the two speeds are different, it then follows that the frequency in air times the wavelength in air, that is, the speed in air, is not equal to the frequency in glass times the wavelength in glass, which is the speed in glass. However, as we’ve already seen, the frequency on both sides of the boundary is the same. So 𝑓 π‘Ž equals 𝑓 𝑔. This being the case, the only way the inequality could possibly hold is if πœ† π‘Ž is not equal to πœ† 𝑔.

Thus, although we haven’t shown on independent physical grounds that the wavelength must be different, we have shown that the frequency is being the same is consistent with the wavelength being different, confirming that (c) is the right answer.

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