### Video Transcript

In this video, our topic is the
critical angle for total internal reflection. As weβll see, this angle shows us
how light can approach a boundary but not cross that boundary. This leads to an important optical
effect called total internal reflection.

As we get started, letβs recall a
few things about a ray of light being incident on a boundary. One of the first things we can say
is that what makes an optical boundary what it is is the fact that the index of
refraction on either side of the boundary is different. So here we have these two indices,
where π one and π two are not equal to each other.

Another thing we can recall is that
our ray of light approaches the boundary at some angle. And we measure this angle relative
to a line drawn perpendicular to the surface where the ray of light reaches it. Itβs common to call this angle, at
which a ray of light is incident on a boundary, the angle of incidence and represent
it using π sub π.

Coming in at this angle, if our ray
wasnβt bent or refracted at all as it crosses this boundary, then its path would
look like this dashed line. But we know this is not the path
the ray of light follows because π one is different from π two. We can say that this boundary line
separates optically different materials. Because of that, if and when our
ray of light crosses over into the material with refractive index π two, it will
refract. It will bend off of this dashed
line that weβve drawn in.

Itβs worth pointing out that, in
general, some of the light in our incoming ray will be reflected off this
interface. But for now, weβre focusing only on
the part that goes into the new material and gets refracted. There are two different ways that
the refracted ray can bend. One way is for the ray to bend
closer to this normal line here. This kind of bending takes place if
the index of refraction of the material our light started in is less than the index
of refraction of the material it moves into. So if π one is less than π two,
we say that our ray bends toward the normal.

And then the second way our ray of
light could bend is away from this normal line. This happens when π one is greater
than π two. In other words, the index of
refraction of the material the ray is leaving is greater than the refractive index
of the material itβs entering.

Now, letβs say we indeed do have a situation where π one is greater than π two. So our ray of light really does
bend like this. And now we can identify another
angle, this one, and recall that this is called the angle of refraction and often
symbolized π sub π. And as a last bit of background,
letβs recall that thereβs a mathematical relationship that connects these variables:
π sub π, π sub π, π one, and π two. That relationship is called Snellβs
law. It tells us that the index of
refraction of the material our ray begins in multiplied by the sine of the angle of
incidence is equal to the index of refraction of the material the ray travels into
times the sine of the angle of refraction.

Now, keeping in mind that the
refractive index of the material our ray starts in is greater than the index of
refraction of the material it moves into, letβs say that we start to change the
angle of incidence of our incoming beam. Specifically, letβs say we start to
increase it. So whereas before we had an angle
of incidence like this, now letβs say that this is our greater angle of
incidence.

Doing this will change the way the
ray of light is refracted as it crosses the boundary. In particular, increasing angle of
incidence also increases angle of refraction. So our new ray may be refracted to
look like this.

Now that we have these new angles
of incidence and refraction, letβs say we increase them even more. Now, this is our new angle of
incidence, and this is our new angle of refraction. And we can see that weβre getting
pretty close to a limit with our refracted angle. To reach that limit, say that we
once again increase our rayβs angle of incidence and that, at this incidence angle,
our refracted ray moves like this. That is, it moves along the
boundary between the materials of our interface.

When this happens, when the angle
of refraction of a beam is 90 degrees, then weβve reached the limit of our incoming
ray actually leaving the first material and entering the second material. At this angle of incidence, it
doesnβt really do either. Rather, it moves along the boundary
between the two. In recognition of reaching this
limit, we give this incidence angle a special name. We call it the critical angle, and
we often represent it using π sub π.

Now that we know this angle exists,
letβs see what happens when we insert it for π sub π in our Snellβs law
equation. When we do this, on the left-hand
side, we have π one times the sin of π sub π. And on the right-hand side, we have
π two times the sine of the angle of refraction, corresponding to the critical
angle of incidence. From our diagram, we can see that
that angle is 90 degrees.

When a ray of light comes into an
interface at the critical angle, the angle of refraction is always 90 degrees. Knowing this, we can recall that
the sin of 90 degrees is equal to one. And that means that the right-hand
side of our equation simplifies to π two. If we then divide both sides of the
equation by π one, then that factor cancels out on the left. And lastly, if we apply the inverse
sine or arc sine function to both sides of the equation, then the application of the
arc sine in the left nullifies the application of the sine function. And so we are left with this
equation here.

If we know the indices of
refraction of both materials on either side of an interface, then we can use that
information to solve for the critical angle of incidence. An important thing to keep in mind
in all this is that weβve required that π one, the index of refraction on this side
of our interface, is greater than π two, the index of refraction on this side. If that wasnβt so, if say π one
was equal to π two or π one was less than π two, then there would be no critical
angle π sub π. It simply wouldnβt exist. That is, if, for example, π two
were less than π one, then thereβs no angle of incidence π sub π we could use
that would lead to a refracted ray being bent in an angle of 90 degrees.

So having a critical angle at all
is only possible if our ray is moving from a higher to a lower index of
refraction. This can be tricky to remember. But thankfully, the form of our
equation for the critical angle helps us. If we forgot that π one must be
greater than π two for π sub π to exist, and say we plugged in an π two that was
greater than π one. If we try to evaluate the inverse
sine of a number larger than one on our calculator or computer, then our device
would either return an error message or an imaginary number. Either result would indicate that
no critical angle exists for the inputs as weβve entered them.

Now, one reason this angle π sub
π is so helpful is that it tells us the smallest angle of incidence required for a
ray of light not to leave the material that it starts out in. At any angle of incidence greater
than π sub π, like this angle of incidence here, our incoming ray will follow a
path where, instead of some of the ray being refracted into the material, like weβve
been seeing, all of it is reflected off its surface. When this happens, we say that a
ray is totally internally reflected. That is, it hasnβt left the
material it started out in.

This phenomenon of total internal
reflection comes in very useful when we want to transmit optical signals over long
distances. When we do this, itβs common to use
whatβs called a fiber optic cable. This cable is set up so that it has
a relatively high index of refraction compared to its surroundings. Which means that when light enters
the cable and then it encounters an interface, instead of being refracted out of the
cable, itβs reflected back into it. And then the next time it reaches
an interface, the same thing happens. The ray of light is reflected back
internally. This takes place over and over
until the ray reaches the other end of the optical cable.

A technology like this requires
knowledge of the critical angle of light as it reaches each one of these boundaries
within the optical fiber. So long as the angle of incidence
is always greater than or equal to the critical angle, the light will remain within
the fiber, even as the fiber bends.

Knowing all this about the critical
angle, letβs get some practice now through an example.

Which of the following formulas
correctly shows the relation between the critical angle for total internal
reflection π sub π for a light ray, the refractive index π sub π of the
substance the light is propagating in, and the refractive index π sub π of the
substance when the light is reflected from its surface? (a) sin of π sub π is equal to π
sub π divided by π sub π. (b) sin of π sub π is equal to π
sub π divided by π sub π. (c) sin of π sub π is equal to
the sin of π sub π over π sub π. (d) π sub π is equal to π sub π
over π sub π. (e) π sub π is equal to sin of π
sub π over π sub π.

Okay, in all these answer options,
we see possible relations between these three variables. π sub π, the critical angle of
incidence; π sub π, the index of refraction of the substance a ray of light starts
in; and π sub π, the refractive index of the substance that the ray of light
reflects off of.

We can begin by making a sketch of
whatβs taking place here physically. Weβre told that we have a
surface. And letβs say that this is our
surface. And itβs there because on one side
of it, say above it, the index of refraction of that material is π sub π. While below this boundary, the
index of refraction of that material is called π sub π.

Our problem statement tells us that
π sub π is the refractive index of the substance that a ray of light is initially
propagating in. So letβs draw that ray of light
like this. And then hereβs what we know. If the angle of incidence of this
incoming ray of light is π sub π, the critical angle, then in that case when this
ray reaches the surface, the boundary between these two materials, it will travel
along that surface. In other words, it will neither
refract into this material nor reflect back into this one. Rather, it moves right along the
boundary between the two. If we consider the boundary that a
ray of light moves along and the line normal to that boundary, then we can see that
this ray is at 90 degrees to that normal line.

Now, even though this ray of light
is in some sense not refracting, in this limiting case, we can nonetheless say that
the angle of refraction is equal to 90 degrees. Indeed, this is the condition for
the angle of incidence to be the critical angle π sub π. So thatβs whatβs taking place in
this situation. We have these two refractive
indices and a light ray approaching the boundary between these materials at the
critical angle.

Knowing all this though, we still
need to choose from among these five candidates for the correct mathematical
relationship between these three variables. Hereβs how we can get started doing
that. We can recall an optical law called
Snellβs law. This law says that if we have a ray
of light here incident on a boundary between two optically different materials. Then the angle of incidence π sub
π of that ray and the angle of refraction π sub π, along with the indices of
refraction of the two materials, are related according to this equation. π sub π times the sine of the
angle of incidence is equal to π sub π times the sine of the angle of
refraction.

Now, this law is generally true,
but itβs also helpful to us in the specific case when the angle of incidence is
equal to the critical angle π sub π. When that happens, when π sub π
is equal to π sub π, then we can see from our diagram here that π sub π is equal
to 90 degrees. So we have these two substitutions
we can make into the variables in Snellβs law. If we write a version of Snellβs
law using these values, then looking at the right-hand side, we see we can do a bit
of simplification. The sin of 90 degrees is equal to
one, which means we can write the equation this way.

And next, if we divide both sides
of this equation by π sub π, the refractive index of the substance the light is
propagating in, then π sub π cancels on the left-hand side. And that leaves us with this
expression. The sine of the critical angle π
sub π is equal to π sub π divided by π sub π. We can now take this expression and
compare it against our five answer options. And we see right away that itβs a
match for option (a). So by using Snellβs law and then
applying the specific conditions implied by a critical angle of incidence, we found
that the sin of π sub π is equal to π sub π divided by π sub π.

Letβs summarize now what weβve
learned about critical angle for total internal reflection. In this lesson, we saw that when a
ray of light incident on a boundary is refracted at 90 degrees, then its angle of
incidence is called the critical angle. We learned further that, for a
critical angle to exist, π sub π, the index of refraction of the material the
light ray begins in, must be greater than π sub π, the index of the material on
the other side of the boundary. When this condition is met, we saw
that the sine of the critical angle, represented π sub π, is equal to the ratio π
sub π divided by π sub π. And finally, we saw that light
reaching a boundary at an angle equal to or greater than the critical angle is
totally internally reflected.