# Video: Calculating the Refractive Index of a Material from the Incident and Refracted Angles

A ray of light traveling in oil is incident on a flat boundary with air, hitting the boundary at an angle of 45 degrees from the line normal to the boundary. The refracted ray in the air travels at an angle of 58 degrees to the line normal to the boundary. What is the refractive index of the oil? Give your answer to one decimal place.

04:09

### Video Transcript

A ray of light traveling in oil is incident on a flat boundary with air, hitting the boundary at an angle of 45 degrees from the line normal to the boundary. The refracted ray in the air travels at an angle of 58 degrees to the line normal to the boundary. What is the refractive index of the oil? Give your answer to one decimal place.

Alright, so in this scenario, we start out with a ray of light that’s traveling in oil. And we’re told that this ray is incident on a flat boundary with air. So let’s say that this is our boundary, where over here on the left, we have oil. And then, over on the right, we have air. We’re told that this ray coming from the side in the oil hits the boundary at an angle of 45 degrees from the line normal to the boundary.

So if this is our ray coming in and reaching this boundary, then at the point where our ray hits this boundary, if we were to draw a normal line, that is, a line perpendicular to the boundary surface. Then we’re told that the angle between that normal line and our ray is 45 degrees. And then, when our ray reaches this boundary, it refracts. That means it bends. We’re told that it does so, so that when the ray moves into the air, it’s traveling at an angle of 58 degrees to this normal line. And that would look something like this. We can see that as the ray has gone across this boundary, it’s moved away from the normal line.

Knowing all this, we want to solve for the refractive index of the oil, the medium the ray was originally in. Now, whenever we have a ray of light going from one material across a boundary into another, we can remember a law called Snell’s law, which describes this situation. Snell’s law has to do with the angle at which a ray is incident on a boundary, the angle at which it leaves that boundary, called the angle of refraction, and also with the refractive indices of the materials on either side of that boundary. In our case, those materials are oil and air.

So, in general, if we have a ray of light moving through a material with an index of refraction called 𝑛 sub one and that ray crosses over a boundary into a material with an index of refraction 𝑛 sub two. And if the angle between the line normal to that boundary and the incoming ray, we call 𝜃 sub 𝑖, and the angle between that normal line and the outgoing ray, we call 𝜃 sub 𝑟, the angle of refraction. Then Snell’s law says that 𝑛 one, the initial index of refraction times the sin of 𝜃 sub 𝑖, the angle of incidence, is equal to 𝑛 two times the sin of 𝜃 sub 𝑟, the angle of refraction.

Now, if we apply Snell’s law to our particular scenario of our ray starting out in oil and going into air, then we can write it this way. We can say that 𝑛 sub 𝑜, the index of refraction of oil, multiplied by the sin of 45 degrees because 45 degrees is the angle of incidence is equal to 𝑛 sub 𝑎, the index of refraction of air, times the sin of 58 degrees, where 58 degrees is the angle of refraction. In this equation, it’s 𝑛 sub 𝑜 that we want to solve for, the refractive index of the oil. In order to do this, we’ll need to know 𝑛 sub 𝑎, the index of refraction of air.

When we talk about refractive index, one thing that’s true about air is that it’s approximately like a vacuum. That is, it has very little effect on the speed of light as it moves through the medium. The refractive index of a pure vacuum is exactly one. And for our purposes, we can treat the refractive index of air as having the same value. This means we can replace 𝑛 sub 𝑎 in our Snell’s law equation with just one. That’s a good approximate value for air’s refractive index.

This means that, to solve for 𝑛 sub 𝑜, the refractive index of the oil, we just need to divide both sides of the equation by the sin of 45 degrees, cancelling that term out on the left side. The refractive index of the oil equals the sin of 58 degrees divided by the sin of 45 degrees. Entering that expression on our calculator, we find a result of 1.19932 and so on. But remember, we are to give our answer to one decimal place. When we look at the value in our first decimal place, a one, we’ll find out whether that stays the same or rounds up by looking at the next value, a nine. Because nine is greater than or equal to five, the one rounds up to a two.

And our final answer then is 1.2. That’s the refractive index of the oil to one decimal place.