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We know from the last few videos

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we have light exiting a slow medium. Let's say I have light ray exiting a slow medium there

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Let me draw. This is its incident angle right over there

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Though it's not the true mechanics of light, you can imagine a car was coming from

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a slow medium to a fast medium; it was going from the mud to the road

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If the car was moving in the direction of this ray, the left tires would get out of the mud

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before the right tires and they are going to be able to travel faster

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So this will move the direction of the car to the right

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So the car will travel in this direction, like that

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where this angle right over here is the angle of refraction

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This is a slower medium than that. This is the fast minimum over here

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We get theta 2 is going to be greater than theta 1

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What I want to figure out in this video is

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is there some angle depending on the two substances that the light travels in

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where if this angle is big enough--because we know that this angle is always

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is always larger than this angle

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that the refraction angle is always bigger than the incident angle

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moving from a slow to a fast medium

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Is there some angle--if I approach it right over here

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Let's call this angle theta 3

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Is there some angle theta 3

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where that is large enough that the reflected angle is going to be 90 degrees

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if that light is actually never going to escape into the fast medium?

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And if I had a incident angle larger than theta 3, like that

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So whatever that is, the light won't actually even travel along the surface

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it definitely won't escape. It won't even travel on surface. It will actually reflect back

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So you actually have something called total internal reflection

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To figure that out, we need to figure out at what angle theta three

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do we have of a fraction angle of 90 degrees?

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That incident angle is going to be called our critical angle

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Anything larger than that will actually have no refraction

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It's actually not going to escape the slow medium

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It's just going to reflect at the boundary back into the slow medium

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Let's try to figure that out and I'll do it with an actual example

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So let's say I have water. This is water

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It has an index of refraction of 1.33

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And let's say I have air up here

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And air is pretty darn close to a vacuum

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And we saw this index of refraction 1.00029 or whatever

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Let's just for sake of simplicity say its index of refraction 1.00

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For light that's coming out of the water

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I want to find some critical angle. I'll call it theta critical

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and so if I have any incident angle less than this critical angle, I'll escape

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At that critical angle, I just kind of travel at the surface

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Anything larger than that critical angle, I'll actually have total internal reflection

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Let's think about what this theta, this critical angle could be

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So I'll break out Snell's Law again

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We have the index of refraction of the water

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1.33 times the sine of our critical angle

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is going to be equal to the index of refraction of the air which is just one

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times the sine of this refraction angle, which is 90 degrees

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Now what is the sine of 90 degrees? To figure that out, you need to think about the unit circle

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You can't just do the soh-cah-toa

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This is why the unit circle definition is useful

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Think of the unit circle

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You go 90 degrees. We are now here on the unit circle

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And the sine is the y coordinate. On a unit circle, that is 1

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So the y coordinate is 1. So this right over here is going to be 1

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So to figure this out, we can divide both sides by 1.33

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So we get the sine of our critical angle is going to be equal to be 1 over 1.33

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If you want to generalize it, this is going to be the index of refraction--

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this right here is the index of refraction of the faster medium

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That right there we can call that index of refraction of the faster medium

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This right here is the index of refraction of the slower medium. So it's ns

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Because the sine of 90 degrees is always going to simplify to 1

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when you're finding that critical angle

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So I'll just keep solving before we get our calculator out

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We take the inverse sine of both sides

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And we get our critical angle. It's going to be the inverse sine 1 / 1.33

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Let's get our handy TI-85 out again

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We just want to find the inverse sign of 1 / 1.33

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which tells us if we have light leaving water at an incident angle of more than 48.8 degrees

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it actually won't even be able to refract; it won't be able to escape into the air

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It's actually going to reflect at that boundary

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If you have angles less than 48.8 degrees, it will refract

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So if you have an angle right over there

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it will be able to escape and reflect a little bit

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And then right at 48.8, right at that critical angle

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you're gonna have refraction angle of 90 degrees

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or really just travel at the surface of water

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And this is actually how fiber-optic cables work. Fiber-optic cables are just--

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You can view them as glass pipes

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And the light is traveling and the incident angles are so large here

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that the light would just keep reflecting within the fiber-optic

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So this is the light ray

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If they travel at larger than the critical angle

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so instead of escaping into the surrounding air or whatever

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it'll keep reflecting within the glass tube

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allowing that light information to actual travel

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Anyway, hopefully you found that reasonably interesting

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Subtitles by Isaac@RwmOne : youtube.com/RwmOne