## Refraction

### Transcript

Though with mirrors we were concerned with reflection for lenses however, we look at refraction. Refraction is governed by Snell's Law, which says the N one Sin data one equals N two Sin data two. Now N is the index of refraction that is given by the speed of light in the vacuum divided by the speed of light in the material of question.

Now we can draw this up, we say if we've got two materials, one that's an index refraction n1, the other n2. A light ray passing through it will bend if these two indices are different. And the angles cut by the light ray will bend in proportion such that it fulfills Snell's law. Now, for this particular example which is the greater index of refraction, n2 or n1?

Well, if we look at Snell's law, if theta, and consequently sin theta, goes up, the index of refraction goes down. So far a small theta we would expect a larger index of refraction. In this case, we have a large theta one, meaning we would expect n1 to be less than n2. So in general, we say if we pass from one material into another with a greater index refraction Y will bend towards the normal.

When light is refracted, what's changing? Frequency, wavelength, velocity. Both the velocity and the wavelength actually change the frequency does not. What's happening is we have a transverse wave like a light ray, as it enters a new material, it slows down. Its forward velocity.

Its frequency or the up-down velocity is the same. So it moves up and down at the same rate but it can't go forward as quickly. So the velocity slows, which consequently makes the wavelength also shorter. Since we know that n1 = c over v1 and n2 = c over v2. We could rearrange this to say, n1v1 equals C, n2v2 also equals C.

So n1 times v1 equals n2 times v2, meaning that the higher the index of refraction, the slower the speed. The slower the speed, the more cramped the wavelength. Continuing on refraction, let's draw the path of the light ray passing through the materials at right. So we've got a light ray.

As it goes from and equals n = 1, n = 2, we would expect it to bend towards the normal. And then as it goes from two to one half to bend away. Now, is there anything wrong with these particular indexes of refraction? You should note that n = 1/2 shouldn't occur how come, N = C/V. C is the speed of light in the vacuum if n equals one half, we're saying that v equals twice the speed of light.

Which means that whatever this material was, it was able to make light go faster than the speed of light. So make sure that your n values are greater than or equal to one. Connect stop is total internal reflection. If we have a light ray that's passing from one material to the next. If it's a very sharp angle it will simply pierce the boundary and refract.

If it's a very wide angle you could reflect back in. So somewhere in between, is an angle such that the scheme is the surface when it hits the boundary. This is called the critical angle. And this angle we can solve for by saying n1 sin theta one equals n2 sin theta two.

This theta two happens to be 90 degrees because this is skimming the boundary. The sin of 90 equals 1, so we get n1 sin theta one equals n2. Or the sin of theta 1 = n2/n1. And this theta 1 becomes our critical angle. And if this angle is bigger than the critical angle, everything is reflected.

So we can write this critical angle equals the arc sign of the ratio of the two indices of refraction. So if we have an angle that's greater than the critical angle, we will reflect. If we have an angle that's less than the critical angle, we refract.

Read full transcriptNow we can draw this up, we say if we've got two materials, one that's an index refraction n1, the other n2. A light ray passing through it will bend if these two indices are different. And the angles cut by the light ray will bend in proportion such that it fulfills Snell's law. Now, for this particular example which is the greater index of refraction, n2 or n1?

Well, if we look at Snell's law, if theta, and consequently sin theta, goes up, the index of refraction goes down. So far a small theta we would expect a larger index of refraction. In this case, we have a large theta one, meaning we would expect n1 to be less than n2. So in general, we say if we pass from one material into another with a greater index refraction Y will bend towards the normal.

When light is refracted, what's changing? Frequency, wavelength, velocity. Both the velocity and the wavelength actually change the frequency does not. What's happening is we have a transverse wave like a light ray, as it enters a new material, it slows down. Its forward velocity.

Its frequency or the up-down velocity is the same. So it moves up and down at the same rate but it can't go forward as quickly. So the velocity slows, which consequently makes the wavelength also shorter. Since we know that n1 = c over v1 and n2 = c over v2. We could rearrange this to say, n1v1 equals C, n2v2 also equals C.

So n1 times v1 equals n2 times v2, meaning that the higher the index of refraction, the slower the speed. The slower the speed, the more cramped the wavelength. Continuing on refraction, let's draw the path of the light ray passing through the materials at right. So we've got a light ray.

As it goes from and equals n = 1, n = 2, we would expect it to bend towards the normal. And then as it goes from two to one half to bend away. Now, is there anything wrong with these particular indexes of refraction? You should note that n = 1/2 shouldn't occur how come, N = C/V. C is the speed of light in the vacuum if n equals one half, we're saying that v equals twice the speed of light.

Which means that whatever this material was, it was able to make light go faster than the speed of light. So make sure that your n values are greater than or equal to one. Connect stop is total internal reflection. If we have a light ray that's passing from one material to the next. If it's a very sharp angle it will simply pierce the boundary and refract.

If it's a very wide angle you could reflect back in. So somewhere in between, is an angle such that the scheme is the surface when it hits the boundary. This is called the critical angle. And this angle we can solve for by saying n1 sin theta one equals n2 sin theta two.

This theta two happens to be 90 degrees because this is skimming the boundary. The sin of 90 equals 1, so we get n1 sin theta one equals n2. Or the sin of theta 1 = n2/n1. And this theta 1 becomes our critical angle. And if this angle is bigger than the critical angle, everything is reflected.

So we can write this critical angle equals the arc sign of the ratio of the two indices of refraction. So if we have an angle that's greater than the critical angle, we will reflect. If we have an angle that's less than the critical angle, we refract.