### Video Transcript

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.