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Current time:0:00Total duration:7:55

Video transcript

we know from the last few videos if we have light exiting a slow medium so let's say so I have light a light ray that's exiting a slow medium right over there and let me draw this is its angle its incident angle right over there we know and the way to visualize it so the way to reason this out if you can run obviously not talking about the true mechanics of light is to imagine if a car was coming from a slow a slow medium to a fast medium if it was going from the mud to the road if the car was moving in the direction of this Ray the left tires were going to get out of the mud before the right tires and they're going to be able to travel faster and so this is going to move the direction of the car to the right so the car is going to travel in a direction in a direction like that where this angle right over here this angle right over there is the angle of refraction and if this is a slower medium than that if this is a fast medium over here we get theta 2 is going to be greater than is going to be greater than theta 1 now what I want to ask or what I want to figure out in this video is is there 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 angles 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 were to approach it right over here if I were to approach at this angle let's just call this this is called this theta 3 is there some angle theta 3 where that is that is large enough that the refracted 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 that so if I had an incident angle larger than theta 3 so let's add an incident angle like that and so whatever that is then it actually that the light won't actually even travel along the surface it definitely won't escape it won't even travel on the surface but it'll actually reflect back so you actually have something called total internal reflection and to figure that out what we need to do is figure out at what angle what angle theta three do we have a refraction angle of ninety degrees at what angle theta three do we have a refraction angle at 90 degrees and then that angle that incident angle is going to be called our critical angle because if you have an angle anything larger than that then you're actually not going to have refraction you're actually not going to escape the slow medium you're just going to reflect at the boundary back in to the slow medium so let's try to figure that out and I'll do it with an actual example so let's say I have water let's say I have 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 you know we saw this index of refraction is one point zero zero zero to nine or whatever but let's just for sake of simplicity say it's index of refraction is one point one point zero zero and what I want to do for light for light that's coming out of the water for light that's coming out of the water let me find another color for light that's coming out of the color I want to find some critical angle some critical angle here I'll call it theta critical where the where the angle of refraction where the angle of refraction is ninety degrees where it will actually never escape so we're this right over here this is ninety degrees and so if I have any angle less any incident angle less than this critical angle I will escape at that critical angle like just kind of travel at the surface and anything larger than that critical angle I will actually reflect I'll actually have total internal reflection reflection so let's think about what this theta this critical angle could be so we'll just break out Snell's log in we have the index of refraction of the water 1.3 3 times the sine of our critical angle times the sine of theta critical is going to be equal to the index of refraction of the air which is just 1 times the sine of this refraction angle so ty the sign the sign of 90 degrees now what is the sine of 90 degrees and the figure that I have to think about the unit circle you can't just do the sohcahtoa because this is actually why the unit circle definition is useful but if you think of the unit circle if you go 90 degrees we are now here on the unit circle and the sine is the y-coordinate so this is a unit circle so that is a 1 that's the one right over here so the y-coordinate when you're right over here is 1 so this right over here is going to be 1 so if we want to figure this out we can just divide both sides by 1.33 so we get the sine we get the sine of our critical angle is going to be equal to well this is just 1 this is just going to be 1 over 1 point 3 3 1 point 3 3 and if you wanted 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 could call that index of refraction of the faster medium and then this right here is the index of refraction of the slower medium I'll call that the slower medium so that right there's n s and because you're always going to find the sine of 90 degrees this is always going to simplify to 1 when you're finding that critical angle so just to keep solving it before we get our calculator out we could take the inverse sine of both sides and we get our critical angle our critical angle is going to be the inverse sine the inverse sine of 1 over 1.1 over 1.3 3 and we can just get our our handy ti-85 out again let's get the ti-85 out and so we just want to find the inverse sine of 1 divided by one point three three and we get 48 point we'll just leave it at 48 point eight degrees so this right here so our critical angle our critical angle is 48 point eight degrees which tells us so this right here is 48 point eight degrees which tells us if we have light leaving water at an incident angle of more than 48 point eight degrees at more than forty eight point eight degrees it actually won't even be able to refract it actually won't be able to escape in the air it's actually just going to it's actually just going to reflect at that boundary if you have angles less than forty eight point eight degrees it will refract so if you have an angle right over there it will be able to escape and refract a little bit and then right at forty eight point eight and this is I've rounded a little bit but right at that critical angle you're just going to travel you're going to have a refraction angle of ninety degrees or really just travel at the surface of the water and this is actually how fiber optic cables work fiber optic cables are just you can just they're just you can almost just do them as kind of glass pipes and the light is traveling that the in the angle of the intensity the incident angles are so large here the light will just keep reflecting within within the fiber optic so this is the light ray they travel it larger than the critical angle so instead of escaping it's get instead of scaping into the surrounding air or whatever it'll just keep reflecting within within the glass tube within the glass tube allowing that light information to actually travel so anyway hopefully you found that reasonably interesting