### 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? Our answer choices are a) π over two, b) three π over two, c) two π over three, d) π, or e) π over three.

In the problem, we are told that the beam of light travels through air into a glass with an index of refraction of three-halves. Index of refraction is represented as a variable π and is defined as the ratio of the speed of light in a vacuum to the speed of light in the 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 index of refraction of air. From the problem, we know that the index of refraction of the 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 known variables, we can say that π subscript a is equal to π, using the variable π subscript a as our wavelength of light in air. We are solving for the wavelength of our beam in 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 the speed of light in air to be the same as the speed of light in a vacuum. From the problem, we are told that the index of refraction for this glass is three-halves.

Recall that the normal line, which is the dotted pink line that weβve drawn in the diagram, is the line that is perpendicular to or normal to the surface. And we can also recall that when light travels from a material of lower refractive index to a material of higher refractive index, it bends towards the normal. Bending towards the normal is a visualization of the light beam slowing down as it enters a medium of higher refractive index from a medium of lower refractive index. Because the speed of light travels fastest in a vacuum, as the index of refraction increases, the speed of light inside the material decreases. Therefore, we can say that the speed of light in air is greater than the speed of light in glass.

Because our equation for index of refraction does not include the wavelength of the beam of light, we need a second equation. π£ equals ππ describes the relationship between the speed of light in a 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 it is travelling in, we can say that the frequency in both air and glass of our light beam stays the same. Because the frequency of light for the light beam is the same in both mediums, then the light beam will have a larger wavelength in the medium that it has the faster speed. Therefore, we can say that the wavelength of the light beam in air will be greater than the wavelength of the light beam in glass.

Now that we have established that the wavelength of the light beam in glass is smaller than the wavelength of the light beam in air, we can eliminate any answer choice in which the wavelength in the glass is greater than or equal to the wavelength in air. Looking back at the problem, we can see that the wavelength in air was π. Therefore, we can cancel out both answer choices b and d. Choice b had a wavelength of light in glass of three π over two or one and a half times the wavelength of the light beam in air. And choice d had a wavelength of the light beam in glass of π, which is the same as 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 in glass, we will need to combine our two equations. Letβs begin by moving around our formula for index of refraction and solving for the speed of light in a vacuum, π. To do this, we multiply both the left side and the right side of the equation by π£, the speed of light in the medium. This leaves us with the equation π times π£ equals π or the index of refraction of the medium times the speed of light in the medium equals the speed of light in a vacuum. Because speed of light in a vacuum is a constant, we can say that the index of refraction of air times the speed of light in air is equal to the index of refraction of glass times the speed of light in glass.

In our problem, we are given the index of refraction for both media and the wavelength of light in both media. However, our equation has the index of refraction for both media and the speed of light in both media. Therefore, we must substitute for the speed of light in both media in terms of the wavelength of light in both media. We know that the speed of light in a medium is equal to the frequency of light times the wavelength in that medium. Our new equation becomes the index of refraction of air times the frequency of light times the wavelength of light in air is equal to the index of refraction of glass times the frequency times the wavelength of light in glass.

Remember that we discussed previously that the frequency of light is dependent on the source and not the medium in which it is travelling. 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. Weβre now left with the equation index of refraction of air times the wavelength of light in air is equal to the index of refraction of glass times the wavelength of light in glass. We are now ready to substitute in our values for our variables.

We can now replace the index of refraction of air with one, the wavelength of light in air with π, the index of refraction of glass with three-halves. And we leave the variable π g as the wavelength of light in glass that we are solving for. 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 one times π or two-thirds π, and the right side of the equation as π g. Our final answer is that the wavelength of the beam in glass is equal to two-thirds π. Looking back at our answer choices, we can see that this is answer choice c.