### Video Transcript

A light ray travelling in water of refractive index 1.3 is incident on the flat surface of a plastic block with a refractive index of 1.6, travelling through the block at an angle of 45 degrees from the line normal to the surface. At what angle from the line normal to the surface does the incident ray strike the block? Answer to the nearest degree.

Okay, lots of information in this question, so let’s underline all the important bits so we don’t miss anything out. So we’ve got a light ray travelling first in water, which has a refractive index of 1.3. And then, it’s incident on the flat surface of a plastic block with a refractive index of 1.6. It travels through the plastic block at angle of 45 degrees from the line normal to the surface. We need to find out the angle from the line normal to the surface that the incident ray strikes the block. We also need to give our answer to the nearest degree.

Now, in questions like this, it’s almost essential that we draw a diagram. This is because it helps us visualise things better as well as label quantities properly and be able to see the entire situation basically. So we’ve got a ray of light travelling in water first and then it meets the flat surface of a plastic block. Let’s say this is the flat surface. So we’ve got water up here and plastic down here.

Now, we don’t know what direction the light ray is travelling in when it’s in the water, but we do know this for when it’s in the plastic. We know that in the plastic the light ray travels at 45 degrees to the line normal to the surface.

So first, let’s draw the line normal to the surface. That’s this line here. And it’s normal to the surface because it’s at 90 degrees to the surface. So we can draw the ray of light in the plastic at 45 degrees to this line. That’s what this looks like. And we can label the angle. There it is, 45 degrees.

Now, we can also extend the line normal to the surface in the other direction so that we can measure the angle of the incident ray. Currently, that we don’t know what this angle is, so let’s just draw any ordinary ray in that we want. Let’s say that it comes in this way. And we can then measure the angle of this ray relative to the line normal to the surface. That’s this angle. Let’s give it a name. let’s call this angle 𝜃 sub 𝑖.

Conventionally, we use the Greek letter 𝜃 to represent an angle and the subscript 𝑖 tells us that it’s the incident ray. But if we’ve given a name to this angle, then we may as well name the other angle that we’ve already labelled. Let’s call the 45-degree angle 𝜃 sub 𝑟. That’s because it’s the refracted angle. In other words, we’re studying refraction here which is when basically a light ray changes direction when it meets an interface, where the refractive index changes.

And speaking of refractive index, we can also label the refractive indices of the water and the plastic. For the water, we’ve been told that the refractive index is 1.3. So we’ll call this 𝑛 sub 𝑖 because traditionally we use the letter 𝑛 to represent a refractive index and the subscript 𝑖 once again tells us that it’s on the incident side. Similarly, 𝑛 sub 𝑟, the refractive index of the plastic, is 1.6. And at this point, we’ve labelled all the quantities we’ve been given in the question onto a diagram.

So let’s go about solving this bad boy. Alright, so we need to find a relationship between 𝜃 sub 𝑖, 𝜃 sub 𝑟, 𝑛 sub 𝑖, and 𝑛 sub 𝑟 such that we know 𝑛 sub 𝑖, 𝑛 sub 𝑟, and 𝜃 sub 𝑟 and we need to find out 𝜃 sub 𝑖. That’s the angle of the incident ray to the line normal to the surface.

The relationship that we’re looking for is known as Snell’s law. Our boy Snell came up with a law that tells us that the product of the refractive index and the sign of the angle made by the rate relative to the line normal to the surface is the same on both sides of the interface. In other words, 𝑛 sub 𝑖 multiplied by sin 𝜃 sub 𝑖 is equal to 𝑛 sub 𝑟 multiplied by sin 𝜃 sub 𝑟.

This is perfect! We’ve already got 𝑛 sub 𝑖, 𝑛 sub 𝑟, and 𝜃 sub 𝑟 and we want to find out 𝜃 sub 𝑖. So we need to rearrange this equation. We can start by dividing both sides of the equation by 𝑛 sub 𝑖. The 𝑛 sub 𝑖’s on the left-hand side cancel, leaving us with sin 𝜃 sub 𝑖 is equal to 𝑛 sub 𝑟 divided by 𝑛 sub 𝑖 multiplied by sin 𝜃 sub 𝑟.

Now, since we’re solving for 𝜃 sub 𝑖, we need to take the inverse sine of both sides of the equation. That looks something like this. And again, the inverse sine cancels with the sine, which leaves us with 𝜃 sub 𝑖 is equal to the inverse sine of 𝑛 sub 𝑟 over 𝑛 sub 𝑖 multiplied by sine 𝜃 sub 𝑟.

And at this point, we can plug in all our numbers. We replace 𝑛 sub 𝑟 with 1.6, 𝑛 sub 𝑖 with 1.3, and 𝜃 sub 𝑟 with 45 degrees. We can then plug this into our calculator, which gives us 𝜃 sub 𝑖 is equal to 60.49 dot dot dot so on and so forth degrees. However, that’s not our final answer because remember the last part of the question tells us that we need to answer to the nearest degree. In other words, the final number in our answer is gonna be this one here because we’re trying to answer to the nearest degree.

Now, in order to find out what happened to this number, we need to look at the one after it. We need to look at this one. It’s this number that will tell us what happens to the zero: whether it rounds up or whether it stays the same. Now, this number here is a four, which is less than five. Therefore, this zero here stays the same. And so- and so our final answer is that the incident ray strikes the block at an angle of 60 degrees to the line normal to the surface to the nearest degree.