Index of Refraction: Study with Video Lessons, Practice Problems & Examples

Refraction is the bending of light as it passes through different materials, which affects its speed. The index of refraction (n) is defined as the ratio of the speed of light in a vacuum (c) to the speed of light in a material (v): $n=\frac{c}{v}$. For example, in water, light slows to approximately $2.2510^-8$ m/s, yielding an index of refraction of 1.33. Generally, n is greater than 1 for materials, while air approximates to 1.0003.

Index of Refraction

Everyone, welcome back. So in this video, we're going to start talking about another phenomenon of light, called refraction. Now we're actually not going to talk about the specifics of refraction in this video. We're going to talk about an important variable that you'll need for that video called the index of refraction, and we'll get to it soon. Alright? So let's just get started. Now remember, when we talk about light, we use this variable c, which stands for the speed of light in a vacuum. This is 3×108 meters per second. The problem is in everyday life, light travels in all different kinds of materials, like air, water, and glass, things like that. What you need to know here is that in all other materials, light always travels slower. It's kind of like how we talked about the speed of sound. The speed of sound differs depending on the material, so does light, and light always travels slower in anything that's not a vacuum. Alright? So that actually kind of brings us to this important variable called the index of refraction. It's given by the variable n over here and basically what it is is it's a ratio. It's a ratio of c, so c is the numerator, to the speed of light in that material. So in other words, it's a ratio of the speed of light in a vacuum to the speed of light in a particular material and that's what n is. So you can actually kind of expand this a little bit and rewrite this as just basically just 3×108 meters per second divided by whatever that speed in that material is. So there's really only 2 variables in this equation, n and v because c is a constant. Alright? Now, actually, I'm going to get back to this point in just a second because we can just jump right into our example. I'm going to show you how this works. So we're told here that when light enters water, it slows to a speed of approximately 2.25×108 meters per second. Now we're going to calculate the index of refraction for water. Alright? So you'll see these tables, very commonly in your textbooks. They'll have all different kinds of variables. You'll never have to memorize them. They'll always be given to you, so don't worry about that. Alright? So we're told the speed of light in water. Which variable is that? Is that n? Is that c? Is that v? Well, hopefully, you guys realized that the c is always just going to be a constant and n is the index of refraction. So, really, they're actually just telling you v. They're telling you the speed of light in that material. So how do we calculate n? n is just equal to c over v. So in other words, it's just equal to 3×108/2.25×108. Now here's just a really quick shortcut when you plug this into your calculator. If these have the same base of 10, then you can kind of just ignore them, and you can only just do, you can just do 3 divided by 2.25. Anyway, what you should get here is you should get 1.33. And if you look in your textbook, that is exactly what the index of refraction of light is. Alright. It's 1.33. Now notice that we got a number that was greater than 1, and that actually kind of brings me back to my point over here. Because what we said here is that light always travels slower in any material, so in other words, if v is always less than c, the way we can see from this equation here is that if you always have a number that's lower than c in the denominator, then that means you're always going to end up with a number that is lower or greater than 1, less than or greater. It's actually going to be greater. So it's going to be greater than 1, and notice how all of these numbers over here are going to be greater than 1. That will always be the case. You'll never see something that's less than 1. Alright? So that's just what you need to know about the index of refraction. The very last point that I have to make here actually has to do with air because it's a very common material that you'll see in problems. So if you look at this table here, it says that the index of refraction is very close to 1.0003. So usually what happens is that in most problems, you can kind of just approximate it and use 1 for the index of refraction when you're talking about air. Alright? That's it for this, folks. Let me know if you have any questions, and I'll see you in the next video.

Diamond has a refractive index of 2.42. How fast would a light ray travel through a diamond?

• A. $2.26\times 10^8$
• B. $2.42\times 10^8$
• C. $1.24\times 10^8$
• D. $2.05\times 10^8$

You turn on one laser in air and shine a second laser through a glass block. How much farther does the light travel in air compared to light traveling in the glass over a period of 2 nanoseconds?

• A. 0.19m
• B. 0.15m
• C. 0.6m
• D. 1.1m

Example 1

Everyone, welcome back. So let's check out this problem here. We have a light ray that's traveling vertically down a frozen lake bed, and it's going to travel through two different materials. The first one is a layer of ice that is 1.5 meters thick. Then after the ice, it crosses into water and travels through another 2 meters of water before hitting the bottom. In other words, if this light ray starts at the surface and goes all the way down to the lake bed, we want to calculate how long it takes for it to actually do that. How long does it take for the travel from the surface all the way down to the lake bed? So we can actually just write that as a Δt because that's a time. Now you can think of this time here because the light travels through two different materials as sort of being made up of two different parts. The first one is Δt_I for ice, and the second one is Δt_W for water. And we know those two times are not going to be the same because the distances involved are different and also the light is going to be traveling through two different materials with two different indices of refraction. So it's definitely not going to be the same Δt. But if I can figure out both of those numbers, if I can figure out Δt_I plus Δt_W, then I could just add these two things together, and that will be my final answer. Alright? So all I have to do is actually just go get what those numbers are. So this is the first number, this is the second one, add them together, and this will be my final answer.

Let's look at the first material. Alright. So let's just look at the ice. So we know that ice is going to travel or the light is going to travel through ice at a certain speed. I'll call this v_I because it's traveling through a material that's not a vacuum or air, so it's going to have some kind of index of refraction. Alright. So this is Δt_I, and in order to figure out a time, I can just use my distance and velocity formula, v equals Δx over Δt. But instead, we're calling them Δx's, we're just going to call them Δy's because they're traveling vertically. Alright? So this is Δy_I for ice, and this will be Δy_W for water. So in order to find a time, I'm going to have to set up my equation that v_I is equal to, Δy_I divided by Δt_I. So I'm looking for the Δt_I, so I could just rearrange this equation. And this just becomes that Δt_I is equal to, if I rearrange this, what happens is these two things trade places. So this is Δy_I divided by v_I. Alright? Now the problem is I actually don't know what v_I is, and I can't just assume that it's 3 x 10⁸ meters per second because that's c. That's the speed of light. So what I have to do is I actually have to go figure this out, this v_I, by relating it to the index of refraction. So I'm going to bring this down here. So n_I is equal to, n_I is equal to c over v_I. So I can rearrange this and say that v_I is equal to 3 x 10⁸ divided by 1.31. That's going to be the index of refraction for ice. What you end up with this, what you end up getting for this is you end up getting a speed of 2.29 x 10⁸ meters per second. So this is the number that I'm actually going to plug back into this formula over here to figure out my Δt_I. So my Δy_I is going to be 1.5 meters. I don't have to convert anything divided by 2.29 x 10⁸. And what I get here out of this, by the way, for Δt_I is I just get a time of this is going to be 6.55 x 10^-9 seconds. Alright. So this is my Δt_I. That's how long it takes for the first one. So now I have to just repeat the same exact step for the water. And go ahead and pause and see if you can actually work it out yourself.

Alright. So let's do the same exact thing for water. Right? So just Δt_W for water. This is going to be that, v_W equals Δy_W divided by Δt_W. Alright? So this just means that Δt_W is just going to be again, you just flip these 2 over. You flip those, you multiply and and and drop the other one down. And this is going to be Δy_W divided by velocity for water. I don't know what that velocity is, so I'm going to have to go to the index of refraction and go get it. So this is just going to be, that v_W is going to end up being c divided by n for water. In other words, it's going to be 3 x 10⁸ divided by 1.33. That's the index of refraction we use for water. It's actually like you're going to have to you know, you can't just go ahead and assume these things are the same. Again, they're slightly different, but you have to go plug in the right number. So that's going to be 1.33. So this v_W ends up being, this is 2.25 x 10⁸ meters per second. This is the number that we plug back into this formula here. So this is going to be 2 divided by, 2.25 x 10⁸. And we'll actually just write this down here, that Δt_W ends up being this is going to be 8.89 x 10^-9 in seconds. Alright? So this is my Δt_W over here, and this is my Δt_I. So actually, I figured out what both of these numbers are, and I can just go plugging them into my final equation. So this final one here is going to be 6.55 x 10^-9, and this one is going to be 8.89 x 10^-9. When you work this out, when you add these two numbers together, what you should get is a grand total of 1.544 x 10^-8, and that's in seconds. So this is our final answer. Alright? So again, just when you have light that's traveling through multiple materials, just break it down. Look at each individual material, solve for those variables, and then work your way down, and you should be able to figure out, the total time for everything. Alright? So that's it for this one. Let me know if you have any questions.

Here’s what students ask on this topic:

What is the index of refraction and how is it calculated?

The index of refraction (n) is a measure of how much light slows down when it enters a material from a vacuum. It is calculated using the formula:

$n=\frac{c}{v}$

where $c$ is the speed of light in a vacuum (approximately 3 × 10⁸ m/s) and $v$ is the speed of light in the material. For example, if light travels at 2.25 × 10⁸ m/s in water, the index of refraction for water is:

$n=\frac{3\times {10}^8}{2.25\times {10}^8}=1.33$

Why is the index of refraction always greater than 1?

The index of refraction (n) is always greater than 1 because light travels slower in any material compared to a vacuum. The formula for the index of refraction is:

$n=\frac{c}{v}$

where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the material. Since $v$ is always less than $c$, the ratio $\frac{c}{v}$ will always be greater than 1. This indicates that light slows down when it enters a material, causing the index of refraction to be greater than 1.

How does the index of refraction affect the bending of light?

The index of refraction affects the bending of light through a phenomenon called refraction. When light passes from one material to another with a different index of refraction, its speed changes, causing the light to bend. The degree of bending is described by Snell's Law:

$n$₁sinθ₁=n₂sinθ₂

where $n$₁ and $n$₂ are the indices of refraction of the two materials, and $\theta$₁ and $\theta$₂ are the angles of incidence and refraction, respectively. A higher index of refraction means light bends more when entering the material.

What is the index of refraction of air and why is it approximated to 1?

The index of refraction of air is approximately 1.0003. However, in most practical problems, it is approximated to 1. This is because the difference between 1 and 1.0003 is very small, and for most calculations, this slight difference does not significantly affect the results. Approximating the index of refraction of air to 1 simplifies calculations without losing much accuracy, making it easier to solve problems involving light traveling through air.

How do you calculate the speed of light in a material given its index of refraction?

To calculate the speed of light in a material given its index of refraction (n), you can rearrange the formula for the index of refraction:

$n=\frac{c}{v}$

Solving for $v$ (the speed of light in the material), we get:

$v=\frac{c}{n}$

where $c$ is the speed of light in a vacuum (approximately 3 × 10⁸ m/s). For example, if the index of refraction of a material is 1.5, the speed of light in that material is:

$v=\frac{3\times {10}^8}{1.5}=2.0\times {10}^8$ m/s.