### Video Transcript

A beam of light has a wavelength π
in air. If the beam passes from air into
glass that has an index of refraction of three-halves, what will the wavelength of
the beam be in the glass? a) π over two, b) three π over two, c) two π over
three, d) π, e) π over three.

In the problem, we are told that a
beam of light travelling through the air enters glass which has an index of
refraction of three-halves. We use the letter π to represent
the index of refraction. We define the index of refraction
as the ratio of the speed of light in a vacuum to the speed of light in a
medium. Because we assume that the speed of
light in air is the same as the speed of light in a vacuum, we can say that the
index of refraction of air is one. We will use the variable π
subscript a to represent the index of refraction for air.

From the problem, we know that the
index of refraction of glass is three-halves. We will use the variable π
subscript g to represent the index of refraction of glass. We also know from the problem that
the wavelength of the light beam in air is π. Including this in our list of known
variables, we can say that the wavelength of light in air, as represented by π
subscript a, is equal to π.

In the question, we are being asked
to solve for the wavelength of the beam in the glass. We can use the variable π
subscript g to represent the wavelength of the light beam in glass. To help us visualize our problem,
we can draw a diagram. From the problem, we know that a
beam of light travels through the air, hits the airβglass boundary, and refracts
into the glass.

Recall that the index of refraction
for air is one because we assume that the speed of light in air is the same as the
speed of light in a vacuum. From the problem, we are told that
the index of refraction for the glass is three-halves. Recall that the normal line, or the
pink dotted line in our diagram, is the line that is 90 degrees, or normal, to the
surface. Also recall that when a light beam
travels from a material of lower refractive index to a material of higher refractive
index, the beam will bend towards the normal.

Bending towards the normal is a
visual representation of the light beam slowing down as it enters a medium of higher
refractive index from a medium of lower refractive index. A decrease in the velocity of light
creates a fraction with a smaller denominator. Because the numerator stays
constant, the value of this expression increases, which means that the value of the
index of refraction, or π, also increases. Therefore, because the index of
refraction of glass is greater than the index of refraction of air, we can say the
velocity of the light beam in glass will be smaller than the velocity of the light
beam in air.

Because our equation for the index
of refraction of light does not contain the wavelength of light, we need a second
equation. The equation π£ equals π times π
describes the relationship between the speed of light in the medium π£, the
frequency of light in the medium π, and the wavelength of light in the medium
π. Because the frequency of a wave is
dependent on the source not the medium itβs travelling in, we can say that the
frequency of the light beam in air is the same as the frequency of the light beam in
glass.

Because the frequency of light is
the same in both air and glass, the medium with the greater speed of light will have
the greater wavelength of light. Therefore, we can say that the
wavelength of the light beam in air is greater than the wavelength of the light beam
in glass. Now that we have established that
the wavelength of the light in glass is smaller than the wavelength of the light in
air, we can eliminate any answer choice that is greater than or equal to the
wavelength of the light in air.

Recalling from the problem that the
wavelength of light in air was π, we can eliminate answer choices b and d. Answer choice b had a wavelength in
glass of three π over two, or one and a half times the wavelength of the light beam
in air. And answer choice d had a
wavelength of the light in glass of π, which is the wavelength of the light beam in
air. This allows us to narrow down our
choices as we begin to solve our problem mathematically.

To solve for the wavelength of the
light beam in glass, we will need to combine our two equations. Letβs begin by moving around our
variables in the equation for the index of refraction and solving for the speed of
light in a vacuum, π. To do this, we multiply both the
left side of the equation and the right side of the equation by π£, the speed of
light in a medium. This will cancel out the π£ on the
right side of the equation.

This leaves us with π£ times π
equal π, speed of light in a medium times the index of refraction of the medium is
equal to the speed of light in a vacuum. We know that the speed of light in
a vacuum is a constant, so the product of the speed of light in air and the index of
refraction of air is equal to π. And the product of the speed of
light in glass and the index of refraction of glass is also equal to π. Therefore, the speed of light in
air times the index of refraction of air is equal to the speed of light in glass
times the index of refraction of glass.

In the problem, we are given the
index of refraction for both media and asked to find, or give in, the wavelength of
light in both media. But in our equation, we have the
speed of light in both media and the index of refraction in both media. Therefore, we must substitute in
for the speed of light in both media in terms of the wavelength. We know that the speed of light in
the medium is equal to the frequency of the light times the wavelength of the light
in that medium. Therefore, we can substitute in π
times π for π£.

After making the substitution, we
get that the frequency of light times the wavelength of light in air times the index
of refraction of air is equal to the frequency of light times the wavelength of
light in glass times the index of refraction of glass. Remember that we discussed
previously that the frequency of light is dependent on the source and not the medium
that itβs travelling in. Therefore, the frequency on the
left side of the equation is the same as the frequency on the right side of the
equation.

If we divide both sides by π, they
end up cancelling out. We are now left with the wavelength
of light in air times the index of refraction of air is equal to the wavelength of
light in glass times the index of refraction of glass. We can now substitute in our values
for our variables. We replace the wavelength of light
in air with π, the index of refraction in air with one. We leave the wavelength of light in
glass as π subscript g, since we are solving for that variable. And we replace the index of
refraction of glass with three-halves.

To isolate the π g, we can
multiply both sides of the equation by two-thirds. This will leave the left side of
the equation as two-thirds times π times one, or two π over three, and the right
side of the equation as π g. Our final answer is that the
wavelength of the light beam in the glass is two-thirds π. Looking back at our answer choices,
we can see that this is answer choice c.