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Studying for a test? Prepare with these 3 lessons on Geometric optics.
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We know from the last few videos we have light exiting a slow medium. Let's say I have light ray exiting a slow medium there Let me draw. This is its incident angle right over there Though it's not the true mechanics of light, you can imagine a car was coming from a slow medium to a fast medium; it was going from the mud to the road If the car was moving in the direction of this ray, the left tires would get out of the mud before the right tires and they are going to be able to travel faster So this will move the direction of the car to the right So the car will travel in this direction, like that where this angle right over here is the angle of refraction This is a slower medium than that. This is the fast minimum over here We get theta 2 is going to be greater than theta 1 What I want to figure out in this video is is there some angle depending on the two substances that the light travels in where if this angle is big enough--because we know that this angle is always is always larger than this angle that the refraction angle is always bigger than the incident angle moving from a slow to a fast medium Is there some angle--if I approach it right over here Let's call this angle theta 3 Is there some angle theta 3 where that is large enough that the reflected angle is going to be 90 degrees if that light is actually never going to escape into the fast medium? And if I had a incident angle larger than theta 3, like that So whatever that is, the light won't actually even travel along the surface it definitely won't escape. It won't even travel on surface. It will actually reflect back So you actually have something called total internal reflection To figure that out, we need to figure out at what angle theta three do we have of a fraction angle of 90 degrees? That incident angle is going to be called our critical angle Anything larger than that will actually have no refraction It's actually not going to escape the slow medium It's just going to reflect at the boundary back into the slow medium Let's try to figure that out and I'll do it with an actual example So let's say I have water. This is water It has an index of refraction of 1.33 And let's say I have air up here And air is pretty darn close to a vacuum And we saw this index of refraction 1.00029 or whatever Let's just for sake of simplicity say its index of refraction 1.00 For light that's coming out of the water I want to find some critical angle. I'll call it theta critical and so if I have any incident angle less than this critical angle, I'll escape At that critical angle, I just kind of travel at the surface Anything larger than that critical angle, I'll actually have total internal reflection Let's think about what this theta, this critical angle could be So I'll break out Snell's Law again We have the index of refraction of the water 1.33 times the sine of our critical angle is going to be equal to the index of refraction of the air which is just one times the sine of this refraction angle, which is 90 degrees Now what is the sine of 90 degrees? To figure that out, you need to think about the unit circle You can't just do the soh-cah-toa This is why the unit circle definition is useful Think of the unit circle You go 90 degrees. We are now here on the unit circle And the sine is the y coordinate. On a unit circle, that is 1 So the y coordinate is 1. So this right over here is going to be 1 So to figure this out, we can divide both sides by 1.33 So we get the sine of our critical angle is going to be equal to be 1 over 1.33 If you want to generalize it, this is going to be the index of refraction-- this right here is the index of refraction of the faster medium That right there we can call that index of refraction of the faster medium This right here is the index of refraction of the slower medium. So it's ns Because the sine of 90 degrees is always going to simplify to 1 when you're finding that critical angle So I'll just keep solving before we get our calculator out We take the inverse sine of both sides And we get our critical angle. It's going to be the inverse sine 1 / 1.33 Let's get our handy TI-85 out again We just want to find the inverse sign of 1 / 1.33 which tells us if we have light leaving water at an incident angle of more than 48.8 degrees it actually won't even be able to refract; it won't be able to escape into the air It's actually going to reflect at that boundary If you have angles less than 48.8 degrees, it will refract So if you have an angle right over there it will be able to escape and reflect a little bit And then right at 48.8, right at that critical angle you're gonna have refraction angle of 90 degrees or really just travel at the surface of water And this is actually how fiber-optic cables work. Fiber-optic cables are just-- You can view them as glass pipes And the light is traveling and the incident angles are so large here that the light would just keep reflecting within the fiber-optic So this is the light ray If they travel at larger than the critical angle so instead of escaping into the surrounding air or whatever it'll keep reflecting within the glass tube allowing that light information to actual travel Anyway, hopefully you found that reasonably interesting Subtitles by Isaac@RwmOne : youtube.com/RwmOne