### Video Transcript

In this video, weโre talking about
the concept in optics known as total internal reflection. As weโll see, total internal
reflection is an idea that relates refracted light, reflected light, and incident
light at an interface. Letโs jump right in.

As we get started, say that we have
an interface between two different materials. The first material, the one on the
bottom side of the interface, has an index of refraction we can call ๐ sub i. And as a side note, we can recall
that index of refraction for a material is equal to the ratio of the speed of light
in vacuum, ๐, to the speed that light takes on in that particular material. The index of refraction of a
material then is a measure of just how much light slows down when it enters that
material.

So on the underside of our
interface, we have a material with index of refraction weโre calling ๐ sub i. And weโll say that, above that
interface, we have a different index, ๐ sub f. And letโs say that, in addition to
this setup, so far we also have a ray of light which is incident on this
interface. Now we ask a question. What will happen to this ray when
it reaches this point?

Well, really, there are two
things. And to help us see what they are
more clearly, we can sketch in this dashed line, which is called the normal line,
thatโs perpendicular to this interface. When this ray of light reaches this
boundary between these two materials, some of the light will be reflected off this
interface. That reflected light ray will look
a bit like this.

And if we call the angle of our
incident light ray measured from the normal line to that ray ๐ sub i, the angle of
incidence. And say we want to relate that
angle of incidence to what we can call the angle of reflection. That angle is measured in a similar
way starting from the normal line. And we can refer to it as ๐ sub
r.

At this point, we can recall whatโs
called the law of reflection, which tells us, using the symbology weโve adopted,
that ๐ sub i, the angle of incidence, is equal to ๐ sub r, the reflected
angle. In other words, these two angles
weโve highlighted here are equal. So thatโs one thing that will
happen to this light beam. Part of it will be reflected back
from the interface. But in most cases, not all of it
will be reflected. Some will be refracted, that is,
transmitted into this second optical material, identified with the index of
refraction ๐ sub f.

In order to show that happening,
weโll need to know something about ๐ sub f compared to ๐ sub i. In particular, weโll need to know
which one of these two values is greater. Letโs say that, in this case, ๐
sub f is less than ๐ sub i. In other words, our ray is going
from an area of higher index of refraction to lower. When this happens, the refracted
portion of this ray bends away from the normal line drawn perpendicular to our
interface. We can also label this angle from
the normal line to the refracted ray. And letโs say that we call it ๐
sub f.

The way we figure out ๐ sub f
given ๐ sub i as well as the index of refraction, ๐ sub i, and index of
refraction, ๐ sub f, is through a law called Snellโs law. This law tells us that the initial
index of refraction, ๐ sub i, multiplied by the sin of the angle of incidence, ๐
sub i, is equal to ๐ sub f times the sin of ๐ sub f, the refraction angle. In other words, given an angle of
incidence as well as an initial and final index of refraction, we can solve for ๐
sub r, the angle of reflection, and ๐ sub f, the angle of refraction.

Given this setup, we may start to
wonder what happens if we increase this angle of incidence. In other words, say that our new
angle of incidence, ๐ sub i, looked like this. If we change ๐ sub i, we can see
that, based on the law of reflection as well as Snellโs law, that ๐ sub r and ๐
sub f will change too. And based on those two laws, hereโs
what our new reflected and refracted rays might look like.

Now letโs imagine that we keep
going in this direction. We keep extending or increasing
this angle of incidence of our ray. Each time we do this and then
recalculate ๐ sub r and ๐ sub f, we start to notice something interesting. As our incident angle, ๐ sub i,
gets bigger and bigger, our refracted angle, ๐ sub f, is even larger than ๐ sub
i. And in fact, ๐ sub i wouldnโt have
to get too much bigger than what weโve shown in order for ๐ sub f, the refracted
angle, to go all the way to 90 degrees. And indeed, there is such an
incident angle, ๐ sub i, for which that happens, through which the refracted ray is
at 90 degrees to the normal line.

Now this is a really very
interesting point weโve reached here because it means that, for this angle of
incidence, no light, whether reflected or refracted, is making it into our second
medium identified by ๐ sub f. And notice something else going on
too. 90 degrees, which is currently the
angle of refraction of our refracted ray, is the biggest that angle could ever
possibly be. The reason is that if that angle
ever exceeded 90 degrees, then that means the ray will never actually enter the
second material. It would be almost reflected back
into the first. In that case, itโs no longer a
refracted or bent ray.

In the situation weโre saying now
where the angle of refraction is 90 degrees, none of the light is making it into the
second material. Under this condition, the angle of
incidence of our incident ray is given a special name. Itโs called the critical angle and
typically symbolized ๐ sub c. This is the angle of incidence for
which the angle of refraction is 90 degrees.

Now like we said, thatโs the most
the angle of refraction can ever possibly be. If we were to increase our angle of
incidence even beyond the critical angle, then in that case we would no longer have
any refracted light. It will all be reflected. At this point, weโve reached a
condition which is called total internal reflection. This is the point at which thereโs
no refracted light because that refracted light ray has reached the 90-degree point
and then been extinguished.

By now, 100 percent of the light
that we send into this interface at the incident angle is reflected back at the
reflected angle. This is the condition known as
total internal reflection. Generally speaking, there are two
necessary conditions for total internal reflection to occur. The first one is a condition we
specified earlier on. We said that the index of
refraction of the material the light ray is trying to get into is less than the
index of refraction of the material itโs already in.

As an example of this, our initial
material could be water and the final material could be air. The index of refraction of water is
greater than that of air. So thatโs one necessary condition
for total internal reflection to occur.

The other necessary condition is
that our angle of incidence, ๐ sub i, needs to be greater than what we called the
critical angle, ๐ sub c. We saw that, at ๐ sub c, at the
critical angle, we still had a refracted ray, granted the ray was being refracted at
90 degrees. But it still existed. Itโs only beyond that critical
angle where we have no refraction at all and we have 100 percent reflection, total
internal reflection.

Now if we bring our angle of
incidence, ๐ sub i, back down to ๐ sub c, the critical angle, we can get a view
into how to calculate this angle. All right, so here we are back at
our critical angle of incidence, ๐ sub c. In addition to our reflected ray,
we have a refracted ray where that angle of refraction is 90 degrees. If we look over to the bottom right
of our screen, we see this equation called Snellโs law. We can use the information in this
equation as well the information in our sketch to give us an expression for this
critical angle, ๐ sub c.

Hereโs how this will work. In Snellโs law, we have the angle
of incidence normally called ๐ sub i. Well, we know that, in this special
case, that angle of incidence is the critical angle, ๐ sub c. So we make that substitution into
the equation. Additionally, on the right-hand
side, we know something about ๐ sub f, the angle of refraction. This angle, according to our
diagram, is 90 degrees. And so we can substitute that value
in for ๐ sub f. Now the sin of 90 degrees is a
value we may have memorized. Itโs equal to one. And this means we can simplify the
right-hand side of this whole expression to read ๐ sub f, the final index of
refraction.

Now our goal here is to get an
expression for ๐ sub c, the critical angle. To help us do that, letโs divide
both sides by ๐ sub i. This has the effect of canceling
that term on the left-hand side. And then as a final step, we can
take the inverse or arc sin of both sides of this equation. When we do that, the entire
left-hand side of this expression simplifies to just ๐ sub c, the critical
angle. And by this point, weโve arrived at
an expression for the critical angle. Itโs equal to the inverse sin of ๐
sub f divided by ๐ sub i. So the critical angle we see just
depends on the ratio of the indices of refraction of the two materials weโre working
with. Thatโs all that matters in
determining this angle. At this point, letโs get some
practice with these ideas through an example.

Which of the following definitions
best describes the critical angle for total internal reflection? A) The critical angle is the angle
at which all incident light at a boundary is reflected. B) The critical angle is the angle
of incidence minus the angle of refraction. C) The critical angle is the angle
of incidence plus the angle of refraction. D) The critical angle is the angle
at which the refracted ray travels along the boundary from which the incident ray is
reflected. And then, last but not least,
option E) the critical angle is the angle of refraction minus the angle of
incidence.

Now in these answer options, weโve
seen a lot of descriptions of angles: angles of refraction, angles of incidence,
angles of reflection. To get some clarity on what these
answer options mean, letโs sketch out a scenario that involves the critical
angle. Say we have an interface here
between two materials. The index of refraction of the
material on top weโll call ๐ sub f, and that on bottom weโll call ๐ sub i. And letโs say further that we have
a ray of light coming up through this first material and reaching the interface
between the two.

Now we donโt need to specify
exactly what ๐ sub f and ๐ sub i are. But it is important that ๐ sub i
be greater than ๐ sub f. The index of refraction of the
material the ray is trying to get into must be less than the index of refraction
that the material is already in. This is a necessary condition for
total internal reflection. In addition to all this, letโs say
that our incoming ray has an angle of incidence that is the critical angle โ weโll
call it ๐ sub c โ that has a very specific meaning in terms of what the refracted
ray looks like.

But before we get to that ray,
letโs look at the reflected ray, the one that bounces off this interface between the
two materials. For this reflected ray, if we call
the angle of reflection ๐ sub r, then by the law of reflection, we know that that
reflection angle, ๐ sub r, is equal to the angle of incidence, which in this case
is the critical angle. So thatโs the reflected ray in this
scenario.

But we also know thereโs a
refracted ray. And because our angle of incidence
is the critical angle, we know that the angle of refraction โ we can call it ๐ sub
f โ is equal to 90 degrees. This is what it means for our
incoming ray to be at the critical angle. So we now have these three angles
marked out: the incident angle, the reflected angle, and the refracted angle. Weโre now ready to evaluate these
five answer options to see which one best describes the critical angle.

We already have answer option E) on
screen. But letโs recall answer options A),
B), C), and D). Answer option A) said that ๐ sub
c, the critical angle, is the angle at which all incident light is reflected. But looking at our diagram, we see
this isnโt so. At the critical angle, we still
have a refracted ray. Itโs at 90 degrees, but it still
exists. So answer option A) isnโt our
choice.

Option B) says that the critical
angle is equal to the angle of incidence minus the angle of refraction. But remember, in the case of the
critical angle, that angle ๐ sub c is equal to ๐ sub i. And the angle ๐ sub f, the angle
of refraction, is 90 degrees. So this equation is essentially
saying that one quantity, ๐ sub c, is equal to another quantity, which weโve said
is identical to ๐ sub c, minus 90 degrees. But mathematically, that canโt be
the case. So B) isnโt our choice either.

And for a similar reason, we wonโt
choose option C). This option says that ๐ sub c, the
critical angle, is equal to the angle of incidence plus the angle of refraction. But again, at ๐ sub c, the
critical angle, the angle of incidence is the critical angle. So this option is saying that a
particular value is equal to itself, in this case, plus 90 degrees. So option C) isnโt our definition
for critical angle either.

Option D) is interesting. It says that the critical angle is
the angle at which the refracted ray travels along the boundary from which the
incident ray is reflected. Now as we look at our sketch, we
see that indeed this is happening. Our refracted ray drawn here is
moving along the boundary at which our incident ray is reflected. So option D) looks like a great
choice.

Letโs just consider option E) to
make sure weโre covering all the bases. This option says that the critical
angle is the angle of refraction minus the angle of incidence. But for the same reason that
options B) and C) were incorrect, option E) isnโt correct either. At the critical angle, the angle of
incidence is equal to the critical angle. So if we subtract this angle from
90 degrees, the angle of refraction, we wonโt get the original angle back. This confirms that option D) is our
best answer.

The critical angle is the angle at
which the refracted ray travels along the boundary from which the incident ray is
reflected.

Letโs summarize now what weโve
learned so far in this lesson about total internal reflection. Weโve seen that total internal
reflection, sometimes abbreviated TIR, is when all the light incident on a boundary
is reflected from it. We saw that total internal
reflection happens when the angle of incidence of a ray exceeds, is greater than,
whatโs called the critical angle, ๐ sub c. At the critical angle, shown here
in this sketch as ๐ sub c, the angle of refraction, shown here as ๐ sub f, is
equal to 90 degrees.

And finally, we saw that if ๐ sub
f is the index of refraction of the material the incident ray is trying to get into
and ๐ sub i is the index of the material itโs already in, then the critical angle,
๐ sub c, is given mathematically as the inverse sin of ๐ sub f over ๐ sub i.