### 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.

Letβs draw a picture of the
situation described in the question. Hereβs 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, weβve labeled this
angle with the symbol π π 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 letβs 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 weβre 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. Letβs call the power transmitted by
the incident ray π zero, by the reflected ray π one, and by the refracted ray π
two. Because thereβs 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 π zero, the power transmitted by the incident ray, is
equal to π one plus π two, the total power transmitted by the reflected and
refracted rays. If we subtract π one from both
sides, we get that π two, the power of the refracted ray, is equal to π zero minus
π 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 π one is greater than zero. However, if π one is greater than
zero, it follows that π zero is not equal to π zero minus π one because π one is
greater than zero. So π zero minus π one must be
less than π zero.

Note that we have a different
expression for π zero minus π one; namely, π zero minus π one is equal to π
two. So letβs substitute this in to π
zero is not equal to π zero minus π one. Doing the substitution, we find π
zero is not equal to π two. And this is exactly the sort of
relationship weβre 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.

Letβs 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 π for the
incident ray on our diagram from before. And now weβve added π π for the
angle between the refracted ray and the line perpendicular to the boundary. To relate these two angles and find
out if theyβre the same or not, weβll use Snellβs law. Snellβs 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. π π, the index of refraction of
air, is approximately one. And π π, 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 weβre talking about the incident ray and the subscript two denote that weβre
talking about the refracted ray, letβs plug in the variables from our diagram into
Snellβs law. We get π π sin π π, the index
of refraction in air times the sin of the angle of the incident ray, is equal to π
π sin π π, the index of refraction in glass times the sin of the angle of the
refracted ray.

Letβs now use the fact that π π
is not equal to π π. If π π is not equal to π π, it
follows that sin π π cannot be equal to sin π π because, otherwise, Snellβs law
would lead to a contradiction. But if sin π π is not equal to
sin π π, then certainly π π is not equal to π π. But if π π is not equal to π π,
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 weβve represented with the symbol π
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. Letβs now plug this into our
inequality π π is not equal to π π. Letting π π be the speed of light
in air and π π 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 letβs multiply both sides by one
over π.

On both sides, π divided by π is
one. And so weβre 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. Weβre left with choices (c)
frequency and (e) wavelength. Letβs 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 weβve 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.

Letβs now examine what happens to
these oscillations at the boundary where the two rays meet. The other fact that weβll need to
know to analyze this situation is that the electric field is continuous across a
boundary. That is, itβs 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, letβs confirm that
wavelength would indeed be different. Weβll do this using the equation
speed of light is equal to frequency times wavelength, where wavelength is
represented with the symbol π. Letβs 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 weβve already seen, the
frequency on both sides of the boundary is the same. So π π equals π π. This being the case, the only way
the inequality could possibly hold is if π π is not equal to π π.

Thus, although we havenβt 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.