Refraction of Light & Snell's Law - Online Tutor, Practice Problems & Exam Prep

Light refraction occurs when light transitions between materials, changing speed and direction. This bending is described by Snell's law, represented as $n_1\sin \left({\theta }_1\right)=n_2\sin \left({\theta }_2\right)$. When light enters a medium with a higher index of refraction, it bends toward the normal, while it bends away when entering a medium with a lower index. Understanding these principles is crucial for solving problems related to light behavior in different materials.

Snell's Law

Welcome back, everyone. We've already talked a lot about light reflection. And in this video, we're going to cover a new concept called light refraction. I'm going to show you exactly what refraction is, and we'll go over the one equation that you need to solve problems, which is called Snell's law. We'll actually go ahead and solve a couple of examples together. So let's get started.

Now when we talked about light reflection, we said that when light comes in and hits a flat shiny surface, it bounces off at the exact same angle that it came in at. In other words, this angle was equal to this angle. But that's actually not the full story because we've also seen from our video on the index of refraction that light can transmit into another material and it will change speed. That's what refraction is about. Whenever light enters a new material at an angle, it changes speed. We've already seen that before, but it also has to change direction. That's what you should think about when you hear the word refraction. Light hits the boundary between two materials and it has to sort of bend at a different angle than the one that it came in at.

Alright. So we actually label these two angles differently. So we can call this one theta one, because it has to do with material 1, and that means that this is also theta one. And that means that this new angle with respect to the normal is equal to theta 2. Alright? And we also have these two different indices of refraction because we have two different materials. This is n1, and this is going to be n2. Now through experimentation, we've actually figured out the relationship between these indices and these angles. And that relationship is this equation here, which is that n1×sin⁡theta1=n2×sin⁡theta2. Your textbooks will refer to this as Snell's law of refraction, named after Snell, the scientist who came up with it.

Basically, what you should know out of this equation from here is that instead of using theta I and theta r, that's what we used for reflection, we're actually just going to use theta 1 and theta 2 for the two different materials that we're dealing with. Alright? This theta 1 here is always going to correspond to your angle of incidence. In other words, it's always just the incoming light, then that should make sense. So in other words, the left side of your equation should always be for the incoming light. The right side of the equation in this theta here is the angle of refraction. And that actually has to do with the outgoing light once it's crossed the boundary into that new material. Alright? That's how you always want to remember this. So basically, just as a note, you should always make sure to use n1 and theta 1 for incident lights. Alright? It'll just help keep track of your variables, and you won't get the wrong answer.

Now that's really it for the equation. Let me go ahead and show you how to use it using these examples. Now in your problems, you're going to have really two possibilities. You either have light that enters a material with a higher index of refraction or a lower one. Those are really only the two possibilities here. So let's take a look at the first one. In this first problem, we have that a ray of light is going to enter water at a 30 degree incident angle. So this is my 30 degrees, and because this is the incident light that's incoming, that's going to be theta 1. We want to find the angle of refraction.

n 1 × sin ⁡ 30 ° = 1.33 × sin ⁡ theta 2

This simplifies down to 0.5, and dividing by 1.33 gives the equation sintheta2=0.51.33. Taking the inverse sine of this fraction gives us theta2=22.1°. This is your angle of refraction. Notice how this number is actually smaller than the 30 degrees of the incident light. So basically, this line is going to get a little bit steeper because remember angles are measured relative to the normal, and this normal here, this angle has to be smaller than the one over here.

So, what we found here is that when you have light that enters a material with a higher index of refraction, the light bends towards the normal, and what we end up with for our theta 2 is a number that is less than theta 1. That's always how these things are going to go.

Now let's go ahead and take a look at the second type of problem where we have light entering material with a lower index of refraction. So here we have a ray of light that's exiting glass into air, same angle, so same 30 degrees over here. This is going to be my theta 1. We know that the index of refraction for glass is 1.46, and for air, it's just n2=1. So, setting up our Snell's law equation:

1.46 × sin ⁡ 30 ° = sin ⁡ theta 2

When you work this out, you get 0.73 equals sin of theta 2, and taking the inverse sin, theta 2 will be 46.9°. In this case, light bends away from the normal and we also get a number that is greater than our initial angle of incidence, which is exactly what we expected. So that's really all there is to light refraction and Snell's law. Let's go ahead and take a look at some practice problems.

A laser pointer emits a ray which enters a quartz crystal at an angle 50° with the normal to the surface of the crystal. The ray bends inside the crystal, making an angle of 30° with the normal. Find the index of refraction of quartz.

• A. 1.00
• B. 0.65
• C. 1.53
• D. 3.77

Example 1

Welcome back, everyone. Let's check out this example problem. So we have a light ray that's going to come in and hit this glass block at some angle. This is going to be my angle of incidence, and it's going to refract. And as it refracts, it's going to go sort of deeper into the glass block until it eventually leaves. And that's actually what we want to calculate. We want to calculate the distance in centimeters below the top of the block that the light ray travels before it exits. Alright? So let's go ahead and set up this problem here. We're now dealing with some kind of refraction problem. This is the angle of incidence. This is my 70 degrees. That's theta 1. And then what happens is after it enters the glass, remember, it's going to be higher end, so it's going to go more normal. So this is air over here. So this is just going to be n equals 1. That's my n one. And this is going to be glass. So this is n₂ = 1.46. I just get this from this table over here. So clearly, it's going to bend more towards the normal. And so therefore, you're going to get a lower angle when you go through to the glass block. So in other words, this theta 2 here is probably going to be a smaller number. Alright? So how does this help us?

Well, if you look here, if we have this angle, can we use that to actually solve for y? Well, if you look at the sort of triangle that we've actually made with the light ray, you can see that the refracted ray sort of makes a triangle with the vertical of the glass block and also the horizontal, the boundary itself. So you look at this little triangle here. Now if we could figure out this angle over here, which I'm going to just call theta x, which gives respect to the x axis. Well, I still can't solve the problem because I have 2 unknowns. I have theta x and I need y. Right? I have y. But there's actually one other thing we know about this triangle. We do know that the ray hits exactly halfway across the glass block, in which the length is 7.5 centimeters. So what that means is if it's exactly halfway, this is just going to be the this distance here that's going to be exactly half of 7.5, which is 3.75 centimeters. And it's okay I leave the measurements in centimeters because that's ultimately what I'm trying to find. Alright? So this is 3.75 centimeters. Remember, to solve a triangle, you need 2 pieces of information, or you either need either need 2 sides, or you need a side and an angle, something like that. So if I could figure out what this theta angle this theta x angle is, then I could definitely use this angle and this distance to figure out this y. Alright? But can I actually figure out what this theta x is? Well, if you take a look at this sort of triangle that we've made, we actually come up with another relationship between theta x and theta 2. Remember, this looks just like, this just is a complimentary angle. This theta x and this theta 2 actually just add up to 90 degrees. So we can actually just set up an equation for this. We can say that theta x is really just equal to 90 minus theta 2. Right? Those things have to add up to 90, so whatever this thing ends up being, this is just going to be 90 minus that. Alright? So this theta x here is 90 minus theta 2. So now can I go and solve for that theta 2? Well, of course, we can, because that's just Snell's law. Right? This Snell's law means that this theta 2 relative to the normal, is going to be related to the angle of incidence and the n one. So we're just going to actually use Snell's law to figure out what this angle is, and then we can go back and figure out what this theta x is. And, actually, we can go ahead and solve our triangle from that. Alright? So that's sort of the game plan here. So I want to set up my Snell's law, my n one sin theta one, n₂ sin theta 2. Alright. So I have what my n one is. Right? That's just the index of refraction for air. Got my angle of incidence. And I also have what the index of refraction for glass is. So I can just go ahead and figure out this theta 2. Alright? So So plugging in some numbers, this is going to be 1 times the sine of 70 equals, and this is going to be 1.46 times the sine of theta 2. Alright? Now when you work this out, what you're going to get here is you're going to get, you're actually just going to get 0.94 equals 0.94 equals 1.46 times the sine of theta 2. You can actually divide this over to the other side and then take the inverse sine. What you should get for theta 2 is the inverse sine of and you can actually do the division or you could just plug it in as a fraction, 0.94 divided by 1.46. And what you should get here is 40 degrees pretty much exactly, like, just to, you know, pretty much round it to 40 degrees. Alright. So this angle here is 40 degrees, and as expected, we got a lower number because we're dealing with a higher index of refraction. Right? Higher and more normal more and more normal. So if this is 40 degrees then, that means we actually can go back and we can solve for our theta x. Remember, Remember, we said that theta x is just 90 minus theta 2. So what this means here is I'm just going to pull this down. This theta x is equal to 90 minus and this is just going to be the 40 that sticks in here. This is going to be 40, and you just get 50 degrees. Alright? So now that we've solved for this theta x, which we now know is 50 degrees here. Now we can go ahead and we can use the angle and this side over here to figure out what my y distance is. Now remember, we're always just going to use pretty much tangent because we'll never gonna know the hypotenuse of the triangles. So let's set up our tangent equation. We say that tangent theta, tangent of theta x is opposite over adjacent. So opposite over adjacent. This is your opposite relative to this angle, and this is your adjacent. Right? So if you set this up, this is just going to be your y value divided by 3.75. And again, we can keep it in centimeters. So now all I have to do is just plug in some numbers here. Right? So I've got the tangent of 50 degrees equals y over 3.75, and I could basically just go ahead and finish this off here by moving this up to the other side. So 3.75 times the tangent of 50 degrees equals the y distance. And if you work this out, what you should get is you should get 4.47 centimeters. And that is your final answer. Alright? So this is going to be 4.47 centimeters. Alright, folks. So that's it for this one. And let me know if you have any questions. Thanks for watching.

A ray of light is incident on a glass pane with an angle of 60°. The light partially reflects and partially refracts. What is the angle θ between the reflected and refracted rays?

• A. 66.4°
• B. 53.6°
• C. 36.4°
• D. 83.4°

Here’s what students ask on this topic:

What is Snell's Law and how is it used in refraction problems?

Snell's Law describes the relationship between the angles of incidence and refraction when light passes between two different media. It is given by the equation:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, $n_1$ and $n_2$ are the indices of refraction for the two media, and ${\theta }_1$ and ${\theta }_2$ are the angles of incidence and refraction, respectively. This law is used to determine the angle at which light will refract when it enters a new medium, given the indices of refraction and the angle of incidence.

How does light behave when it enters a medium with a higher index of refraction?

When light enters a medium with a higher index of refraction, it slows down and bends toward the normal. This means the angle of refraction (${\theta }_2$) is smaller than the angle of incidence (${\theta }_1$). This behavior can be explained using Snell's Law:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Since $n_2$ is greater than $n_1$, ${\theta }_2$ must be smaller to satisfy the equation.

What happens to light when it exits a medium with a higher index of refraction into a lower one?

When light exits a medium with a higher index of refraction into a lower one, it speeds up and bends away from the normal. This means the angle of refraction (${\theta }_2$) is larger than the angle of incidence (${\theta }_1$). Using Snell's Law:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Since $n_1$ is greater than $n_2$, ${\theta }_2$ must be larger to satisfy the equation.

How do you calculate the angle of refraction using Snell's Law?

To calculate the angle of refraction using Snell's Law, you need the indices of refraction for both media and the angle of incidence. The law is given by:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Rearrange the equation to solve for the angle of refraction (${\theta }_2$):

${\theta }_2=\sin -1\left(\frac{n_1\sin {\theta }_1}{n_2}\right)$

Plug in the known values for $n_1$, $n_2$, and ${\theta }_1$ to find ${\theta }_2$.

What is the index of refraction and how does it affect light refraction?

The index of refraction, denoted as $n$, is a measure of how much a material slows down light compared to a vacuum. It is defined as:

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

where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the material. A higher index of refraction means light travels slower in that material, causing it to bend more when entering or exiting the material. This bending is described by Snell's Law.