Now we’re ready to introduce Snell’s law. We said that the incident and reflected angle are the same
which leaves us to have to figure out the transmitted angle. We already mentioned that light will bend
when it goes into a new medium. So, we’re simply asking ourselves: by how much does the light bend as it
goes from one medium with one index of refraction like a vacuum into another medium with a different
index of refraction? This is given by Snell’s law. Snell’s law says that the index of refraction times the
sine of the incident angle will be equal to the new index of refraction times the sine of that transmitted angle.
Going into a denser medium, as you can see just by looking at this equation, will give you a smaller theta.
You can sort of examine this equation. It shouldn’t necessarily be immediately apparent. But just look
at this equation for a bit because Snell’s law is very often used. It’s a very important equation.
Again, it tells us exactly how an incident beam will turn into a transmitted beam with a smaller angle
if it’s going into a denser medium. By the opposite token, if it's going into a sparser or lighter or less dense
medium, the transmitted angle, the angle at which it leaves that medium will instead be bigger.
So, we have these two different behaviors. They’re both given by Snell’s law. Now that we know Snell’s law,
there’s one other important thing we could derive about Snell’s law. Imagine that we have something
like this that we have in the bottom here. A light ray is going from some dense medium, maybe it’s water
into air or maybe it’s air going into a vacuum, going into a lighter medium. As we increase the angle
of incidence, the angle at which the light hits the less dense medium, the transmitted beam is going to
change how much it’s transmitted by in the reflected angle, the transmitted angle that is.
Looking at the Snell’s law for this particular scenario, we can see that we increase the incident angle.
We could ask ourselves, "At what incident angle will the beam now have a 90-degree of reflecting
or transmitting angle?" So, looking at the middle case here, what angle, what incident angle would we need
in order for the transmitted beam to not be transmitted out at an angle anymore but be transmitted
horizontal to or at a perpendicular angle to the new medium? All we have to do is actually plug this in.
So, we take this middle case. We take θ2, the new angle to be 90 degrees. Then we remember
that the sine of 90 degrees is just one. So, we have the n1 times the sine of the incident angle
will be equal to n2, the new medium since the sine of theta transmitted which is 90 degrees is just one.
We can rearrange this equation and try to solve for the incident angle that we need for this effect to occur.
We get something like this. That the sine of the incident angle, the sine of θ1 has to be equal to
the ratio of the indices of refraction of the two media. A few important things about this: First of all,
look at the equation for the sine of this angle. We call this angle the critical angle, the angle at which
the light, as it’s trying to leave this medium, will bend and be horizontal to or perpendicular as you can
see here to the new surface. This is the critical angle. But notice that the sine of the critical angle is equal to
the ratio of these two indices of refraction. But we already know about the sine of theta, that the sine
of an angle can never be greater than one. For this reason, we need for n1, the index of refraction
of the initial medium that it’s starting from to be greater than n2, the index of refraction
of the medium it’s going into because otherwise, we’ll have sine of θ equals some number greater
than one. That can never be the case. For this reason, the initial medium, the medium that the light
is starting in has got to be more dense than the medium it is going into. This is the only way
you can ever get what we call total internal reflection. We call it total internal reflection because this beam,
as it’s trying to leave this dense medium can never get out so long as the angle, this incident angle
is always greater than this number that we found here for the critical angle. For example, this could be
used in fiber optics. If you have a fiber optic cable made of some sort of glass material
and light is traveling through that cable, if it’s trying to get out of that cable and go to a less dense medium
than the glass, it will not be able to as long as we keep the angle shallow enough, as long as we keep
that incident angle beyond this critical angle requirement. Then we can maintain our total
internal reflection. Finally, we have one last topic with this introduction to optics which is to point out
that in fact for different wavelengths of light, there can actually be different indices of refraction
for each wavelength. For example, the index of refraction for red light might be less than the index
of refraction for blue light. By Snell’s law, this means that different colors would bend by different
amounts when they enter a new medium. We can use this effect which we call dispersion by sending
light into some denser medium expecting that light, if it has different components to its color,
maybe white light that has many different colors in its spectrum as you can see here and use
this effect, this dispersion effect of each color having a different index of refraction and a tool
called a prism. You might be familiar with the idea of a prism. You send the light in one side which has
many components to its color. Then because of this dispersion effect, we have all the different
wavelengths separated on the other end, each one at its own frequency. This, in fact, can be used
to analyze light coming in from some source to see what frequency components are present in that
source of light. This summarizes or wraps up our initial intro to optics. We’re going to be using more
ideas of optics when we get to lenses and mirrors especially which is what we’ll do next.
Thanks for listening.