### Video Transcript

A light ray traveling in air is incident on the flat surface of a plastic block, hitting the surface at an angle of 45 degrees from the line normal to it. The refracted ray in the block travels at an angle of 33 degrees to the line normal to the surface. What is the refractive index of the plastic?

Okay, so in this question, we’re dealing with a ray of light initially travelling in air. So, let’s say that this is the ray of light. And we’ve been told that it’s incident on the flat surface of a plastic block. So, let’s say that this is the plastic block. Now additionally, we’ve been told that the ray of light, which is initially travelling in air, is hitting the surface of the plastic block at an angle of 45 degrees from the line normal to the surface. So, if this is the surface of the plastic block, then this is the line normal to that surface because the normal line is at 90 degrees to the surface, perpendicular to the surface. That’s what it means for it to be normal.

Now, we’ve been told that the angle between our ray of light and the line that is normal to the surface of the plastic block, so that’s the angle between our ray and our normal line here, is 45 degrees. Let’s call this angle 𝜃 subscript one. And we know that it’s 45 degrees, so we’ll label it as such. Now, because our ray of light is travelling from air to plastic, in other words, from a medium with a certain refractive index to a medium with a different refractive index, our ray of light is going to refract or change direction in the plastic block. And in fact, we’ve specifically been told that the refracted ray in the block travels at an angle of 33 degrees to the line normal to the surface.

And so, if we imagine that this is the ray of light now travelling through the plastic block, then we can say that the angle between that ray of light and the normal line, which we will call here 𝜃 subscript two, is equal to 33 degrees as we’ve been told in the question. Now, to make things clearer at this point, we can label the two media that the ray of light is travelling through. Initially, it’s travelling through air as we’ve been told in the question, and then it travels into a plastic block. And the important thing here is that air and plastic have different refractive indices. That’s what causes our ray of light to refract when it enters the plastic. We’ve already discussed this because, otherwise, the ray of light would’ve carried on travelling in this direction.

But instead it doesn’t; it refracts. And we need to work out the refractive index of the plastic that results in this specific refraction. Now, to do that, we also first need to remember that the refractive index of air, which we will call 𝑛 subscript one, is approximately equal to one. And for our intents and purposes, we can take the refractive index of air to be exactly one. This is because light travelling through air behaves almost identically to light travelling through a vacuum. And a vacuum has a refractive index of exactly one. And then, we can say that the refractive index of the plastic block, we will call 𝑛 two, and this is exactly what we’re trying to find.

So, the variables that we have to work with are 𝑛 one, the refractive index of the air, 𝑛 two, the refractive index of the plastic block, 𝜃 one, the angle between the ray of light in the air and the line normal to the surface, and 𝜃 two, the angle between the ray of light in the plastic block and the line normal to the surface. It’s at this point we can recall a relationship known as Snell’s law. Snell’s law is a relationship between exactly these four quantities that we’ve just discussed. And it tells us that the refractive index of one material multiplied by the sin of the angle between the ray of light and the line normal to the surface is equal to the refractive index of the material on the other side of the surface multiplied by the sin of the angle between the ray of light in that surface and the line normal to the surface.

And so, in this particular case, as we’ve already labelled them, 𝑛 one is the refractive index of air, 𝜃 one is the angle between the ray of light in air and the line normal to the surface, 𝑛 two is the refractive index of the plastic block, and 𝜃 two is the angle between the ray of light in the plastic block and the line normal to the surface. Which means that if we want to solve for 𝑛 two, the refractive index of the plastic block, we can do this by dividing both sides of the equation by the sin of angle 𝜃 two. Because if we do this, we’ve got a sin of 𝜃 two in the numerator and denominator on the right-hand side. Which means we have sin 𝜃 two divided by sin 𝜃 two, which is just equal to one. And so, what we’re left with is 𝑛 one multiplied by sin 𝜃 one divided by sin 𝜃 two is equal to 𝑛 two.

At this point, we can substitute in the values that we already know on the left-hand side of this equation. But because we’re solving for 𝑛 two, we’ll put 𝑛 two on the left-hand side. And on the right-hand side, we’ll have 𝑛 one — that’s the refractive index of air, which we’ve said to be one — multiplied by the sin of the first angle 𝜃 one, which is 45 degrees, divided by the sin of 𝜃 two, which happens to be 33 degrees. And so, at this point, we can plug this expression on the right-hand side into a calculator, and this gives us a value for 𝑛 two of 1.2983 dot dot dot, so on and so forth. Now, notice that refractive indices are unitless quantities. In other words, we’re not going to have a unit at the end of this number.

And very quickly, the reason for this is that refractive index is defined as the speed of light in a vacuum divided by the speed of light in the material whose refractive index we’re trying to calculate. And so, 𝑛, the refractive index, is equal to a speed divided by a speed. The units of speed in the numerator and denominator cancel, leaving us with a unitless quantity. And we can see that this is obeyed in our expression here as well. We know that the refractive index of air is once again a unitless quantity because it’s a refractive index. And we also know that the sin of an angle does not have a unit divided by the sin of an angle which also does not have a unit.

And so, we find that the refractive index of the plastic block, 𝑛 two, which is what we’re trying to find, is 1.2983 dot dot dot. However, because the information that we’ve been given in the question, specifically the angle 𝜃 one which is 45 degrees and the angle 𝜃 two which is 33 degrees have both been given to us to two significant figures. In the case of 45 degrees, this is the first significant figure, this is the second. And the same can be done for 33 degrees.

This therefore means that we have to give our answer to two significant figures as well. So, here is the first significant figure and here is our second. Now, in order to work out what happens to our second significant figure, we need to look at the next value. This one is a nine, and nine is greater than or equal to five. Which means our second significant figure, our two, is going to round up. It’s going to become a three. And so, we can say that to two significant figures, the refractive index of the plastic making up the plastic block is 1.3.