Refraction of Light & Snell's Law - Video Tutorials & Practice Problems
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1
concept
Snell's Law
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7m
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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 gonna 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. But, 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 all, right. So we actually label these two angles differently. So we, we can call this one theta one because it has to do with material one. 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 two. All right. And we all have these two different indices of refraction because we have two different materials. This is N one and this is gonna be N two. Now, through experimentation, we've actually figured out the relationship between these indices and uh these indexes and these angles. And that relationship is this equation here, which is that N one times the sine of theta one is equal to N two times the sign of theta two, 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 this equation here is that instead of using theta I and theta R, that's what we used for reflection. We're actually just gonna use theta one and theta two for the two different materials that we're dealing with. All right, this theta one here is always going to correspond to your theta or your angle of incidence. In other words, it's always just the incoming lights and 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 data here is the angle of refraction. And that, that actually has to do with the outgoing lights once it's crossed the boundary into that new material. All right, that's how you always want to remember this. So basically, just as a note, you should always make sure to use N one and theta one for incident lights. All right, it'll just help keep track of your variables. 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 gonna have really two possibilities. You either have light that enters a material with a higher index of a fraction 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 gonna 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 gonna be theta one. We wanna find the angle of refraction. Now, the problem is, I don't know what that angle is. So I can't draw it here yet. But basically what happens is that this thing is gonna bend either this way or this way, it's gonna bend off in some direction. And I wanna find that. All right. Now, what other, what else do I know? I know the two materials that are involved. I know that air for N one is just uh equal to one. And I know that the N for water, which I'll call N two is equal to 1.33. So if I want to find the angle, I just have to use my new equation for Snell's Law which remember just says that N one times sine whoops and one times sine theta one equals N two, sine theta two. So this theta two here is my angle of a fraction. That's what I'm trying to solve for. But I have three out of four variables. I know the ends for both of them. And I also have the incident angle and that's usually how these problems are gonna go. You're gonna have three out of four variables and you'll just have to solve for the missing one. All right. So this N one here is just one, this sign of theta is just gonna be sign of 30 and this is gonna be 1.33 times the sign of the two. All right. So just to uh simplify this, this is gonna simplify down to 0.5 and we can also divide over the 1.33. So this is gonna be divided by 1.33 and this equals sign of theta two as one last step here which I just have to get rid of the sign to isolate that theta and I just take the inverse sign. So basically what happens is the two equals the inverse sign. And you could solve for this or you could just plug it in as a fraction 0.5 divided by 1.33. And you've got an angle of 22.1 degrees. So this is your angle of refraction here, notice how this number is actually smaller than the 30 degrees of the incident light. So that means that when you draw your line, this angle here has to actually be closer to the normal than it was over here. So basically, this line is gonna 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 this is 22.1 degrees. That's your angle of refraction. So basically what we found here is that when you have light that enters a material with a higher index of refraction, the light bends, does it bend toward or away from the normal? But what we just saw here is that it actually bends and gets closer to the normal. So light bends towards the normal. And what we end up with for our theta two is we end up with a number that is less than theta one. That's always how these things are gonna go so light bends towards the normal, uh for these kinds of problems. All right. So 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 a fraction. So let me ask you if this is basically just the opposite of this situation over here. Which way do you think light is going to bend? Is it gonna bend toward or away from the normal? Well, I guess that it's just gonna bend away from because it's the opposite of the left situation. And we're gonna end up with a situation where theta two is greater than theta one. But let's go ahead and calculate it just so we can make sure. So here we have a ray of light that's exiting glass into air, same angle. So same 30 degrees over here. This is gonna be my theta one. We know that the index of refraction for glass is 1.46 and for air, it's just N two that equals one. All right. So notice how, what happens is it doesn't matter which one is greater or smaller. The incoming light is always gonna labeled as N one. All right. So let's set up our Snell's law. We have the N one sin, theta one equals N two sine theta two. So we're looking for theta two, we just plug in all the rest of the numbers. So N one is equal to 1.46 times the sign of 30 then N two is just gonna equal one. So that just uh doesn't change anything. All right. So when you work this out, what you're gonna get here is 0.73. So this 0.73 equals sine of theta two. And all you have to do is just take the inverse sign. So this is gonna be that theta two is gonna be the inverse sign of 0.73. And which you'll end up with here is 44 sorry 46.9 degrees. So this is my new angle notice how this number is bigger than the 30 degrees. So that means when I go over to my diagram, I have to draw a an array that's gonna end up being farther away from the normal than this one. So here, what happens is that you end up with a large angle, this is 46.9 degrees. And what we can say here is that light bends away from the normal just as we expected. And we also got a number that is greater than our initial angle of incidence, which is exactly again what we expected. So that's really all there is to the uh the light refraction and Snell's law. Let's go ahead and take a look at some practice problems.
2
Problem
Problem
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
3
example
Example 1
Video duration:
6m
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Welcome back everyone. Let's check out this example problem. So we have a light ray that's gonna come in and hit this glass block at some angle. This is gonna be my angle of incidence and it's gonna refract and as it refracts, it's gonna go sort of deeper into the glass block until it eventually leaves. And that's actually what we wanna calculate. We want to calculate the distance in centimeters below the top of the block that the light ray travels before exits. All right. So let's go ahead and set the set up this problem here. We're now we're dealing with some kind of refraction problem. This is the angle of incidence. This is my 70 degrees. That's theta one. And then happens is after it enters the glass, remember it's gonna be higher N so it's gonna go more normal. So this is air over here. So this is just gonna be N equals one, that's my N one and this is gonna be glass. So this is N two equals 1.46. I just get this from this table over here. So clearly, it's gonna bend more towards the normal. And so therefore, you're gonna get a lower angle when you go over, when you get through the, to the glass block. So in other words, this theta two here is probably gonna be a smaller number. All right. So how does this help us? Well, if you look here, if we have this angle does, can we use it 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 gonna just call theta X, which because in respect to the X axis, well, I still can't solve the problem because I have two 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 gonna be that this distance here is gonna be exactly half of 7.5 which is 3.75 centimeters and it's OK. Leave the leaves, um my measurements in centimeters because that's ultimately what I'm trying to find. All right. So this is 3.75 centimeters. Remember to solve a triangle, you need two pieces of information, you need, you need two sides or you need a side at an angle, something like that. So if I could figure out what this the angle, this the X angle is, then I could definitely use this angle and this distance to figure out this why? All right. But can I actually figure out what this data 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 two. Remember this looks just like um this just is a complementary angle, this theta X and this theta two actually just add up to 90 degrees. So we can actually just set up an equation for this. We could say that theta X is really just equal to 90 minus theta two, right? Those things have to add up to 90. So whatever this thing ends up being, this is just gonna be 90 minus that. All right. So this theta X here is 90 minus theta two. So now can I go and solve for that? Theta two? Well, of course, we can, because that's just Snell's law, right? The Snell's law means that this theta two relative to the normal um is gonna be related to the angle of incidents and the N one. So we're just gonna 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. All right. So that's sort of the game plan here. So I'm gonna set up my Snell's Law, my N one S the one and two sin theta two. All right. So I have what my N one is right? That's just the uh index of fraction for air got my angle of incidence. And I also have what the index of a fraction for glasses. So I can just go ahead and figure out this theta two. All right. So plugging in some numbers, this is gonna be one times the sign of 70 equals and this is gonna be 1.46 times the sign of theta two. All right. Now, when you work this out, what you're gonna get here is you're gonna get um you're actually just gonna get 0.94 equals 0.94 equals 1.46 times the sign of theta two. You can actually divide this over to the other side and then take the inverse sign. What you should get for theta two is the inverse sign of and you can actually do the division or you could just plug it into this 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. All right. So this angle here is 40 degrees. And as expected, we got a lower number because we're dealing with a higher index of a fraction, right, higher end, more normal, more and more normal. So this is 40 degrees. Then that means we actually go, you can go back and we can solve for our theta X. Remember we said that theta X is just 90 minus theta two. So what this means here is, I'm just gonna pull this down, this theta X is equal to 90 minus. And this is just gonna be the 40 that sticks in here. This is gonna be 40 and you just get 50 degrees. All right. So now that we've solved for this theta X, which we know now 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're 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 gonna 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 tangents 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. All right. So this is gonna be 4.47 centimeters. All right, folks. So that's it for this one. And uh let me know if you have any questions. Thanks for watching.
4
Problem
Problem
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°
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