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 a fast medium 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 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 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 a refraction 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
And we get 48.8 degrees.
So this right here, so our critical angle is 48.8 degrees
So this right here is 48.8 degrees
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 refract 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