Name: Physics 53 Polarization (5 of 5) Brewster's Angle
Uploaded: 2024-10-02T09:19:43.000Z
Duration: 6 min 26 s
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Welcome to iLecture Online and here's the last part of the topic, polarization.

In this case we discover that there's a very interesting way in which light can be polarized.

And that light is associated with something called Brewster's angle.

It was discovered that when light shines onto a surface, like a boundary, some of the light will be transmitted, or refracted as we should say, and some of the light will be reflected.

So part of the light will be reflected off the surface, part of the light will be transmitted or refracted through the surface.

And of course, if there's a difference in the index of refraction, if this is N1 and N2, and N2 is larger than N1, we realize that if we then draw the normal to the surface, that the angle of incidence, theta sub 1, is going to be larger than the refracted angle, theta sub 2.

And it turns out that if the light, which is unpolarized as it approaches the boundary, gets reflected in such a way that the angle between the reflected and the refracted angle is equal to 90 degrees, then the light that gets reflected will be completely polarized, just like as if it went through a polarizer.

Now of course in this example you can see that it's not quite 90 degrees.

But if we have an example where the light comes in at just the right angle, so that it reflects in such a way that the refracted angle makes a 90 degree angle with the reflected angle, if this is equal to 90 degrees, then the light that is reflected is polarized.

And so the angle of the incident light is then considered Brewster's angle.

What is the angle that it needs to be, that the inbound light is at relative to the normal of the surface so that the light that's reflected will be polarized?

And of course what we need to know is what n1 and n2 is.

And let's assume that this is air, so that n1 is equal to 1 for air.

And if this is, let's say, water, a smooth water surface, then n2 would be equal to 1.33.

In this particular case, what would Brewster's angle be equal to?

Alright, so remember that the condition of the reflected light and the refracted light, that the angle between them should be 90 degrees.

Now if we continue this normal line right here, and we call this here theta sub 2, and then knowing that this is the incident light, and then this would be the reflected light, theta reflected, which of course has to be equal to theta sub 1, the incident light, which is equal to Brewster's angle.

Alright, so since the distance or the angle from there to there is 180 degrees, if you subtract 90 from that, we then see that theta sub 1 and theta sub 2 added together therefore must be equal to 90 degrees.

So theta sub 1 plus theta sub 2 must equal 90 degrees.

Now if we use Snell's Law, where we can see that n1 sine of theta 1 is equal to n2 sine of theta sub 2, and then realize that theta sub 2 can be written as 90 degrees minus theta sub 1, and then reinsert it into this equation from Snell's Law, we then get n1 sine of theta 1 is equal to n2 times the sine of 90 degrees minus theta sub 1.

And now notice we now have an equation with only theta sub 1 in it.

Of course we still have this 90 degrees there which we want to get rid of.

So how do we get rid of that 90 degrees minus theta sub 1?

Well for that we need to know a trigonometric identity.

We know that the sine of A minus B can be written as the sine of A, cosine of B minus cosine of B, cosine of not B but A, cosine of A, sine of B.

Alright, if we then apply that to what we have over here, we can then say that the sine of 90 degrees minus theta sub 1 is equal to the sine of 90 degrees times the cosine of theta sub 1 minus the cosine of 90 degrees times the sine of theta sub 1.

And of course the cosine of 90 degrees is equal to zero, so this term disappears and the sine of 90 degrees is equal to 1.

So we then can say that the sine of 90 degrees minus theta sub 1 is equal to the cosine of theta sub 1.

So in this Snell's Law here we can replace sine of 90 degrees minus theta sub 1 by cosine of theta sub 1.

So now we have n1 sine of theta 1 is equal to n2 times the cosine of theta sub 1.

And now again we have a simplified equation where the only variable is theta sub 1, only one variable.

So we're going to solve that for theta sub 1.

So we're going to divide both sides by cosine of theta and both sides by n sub 1, which means we now have the sine of theta sub 1 divided by the cosine of theta sub 1 by moving the cosine down here is equal to n sub 2 divided by n sub 1.

And of course the sine divided by the cosine, that's equal to the tangent.

So now we can say that the tangent of theta sub 1 is equal to n2 over n1, which means that theta sub 1 is equal to the arctangent of n2 over n1.

And so this is equal to the arctangent of 1.33 over 1.

Let's find out what that's equal to, 1.33.

Take the arctangent of that and we get 53.06 degrees, 53.06 degrees.

That's the angle required for theta sub 1 to have reflection that will be polarized because at that moment the angle between the reflected light and the refracted light will be equal to 90 degrees and that is then known as Brewster's angle.

So in this case, Brewster's angle is equal to 53.06 degrees and when you shine light at that angle relative to the normal on a smooth water surface, the reflected light will be polarized.

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Physics 53 Polarization (5 of 5) Brewster's Angle

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