Video Transcript
A light ray traveling in air is
incident on the flat surface of a plastic block, hitting the surface at an angle of
45 degrees from the line normal to it. The refracted ray in the block
travels at an angle of 33 degrees to the line normal to the surface. What is the refractive index of the
plastic? Give your answer to one decimal
place.
In this question, we’re dealing
with a ray of light initially traveling in air. The ray is incident on the surface
of a plastic block, and we want to determine the refractive index of the
plastic. We will start by drawing out this
problem.
So let’s begin by saying that this
is the initial ray of light. We’ve been told that this ray is
incident on the flat surface of a plastic block. So let’s say that this is the
plastic block. To make things clear, let’s label
the air and plastic block as follows. Now, additionally, we’ve been told
that this ray of light hits the surface of the plastic block at an angle of 45
degrees from the line normal to the surface. So, if this is the surface of the
plastic block, then this is the line normal to that surface, because the normal line
is at 90 degrees to the surface, perpendicular to the surface. That’s what it means for it to be
normal. And so the angle between the ray
and the normal line here is 45 degrees. Let’s call this angle 𝜃 sub
one.
Now, because this ray of light is
traveling from air to plastic, in other words, from a medium with a certain
refractive index to a medium with a different refractive index, this ray of light is
going to refract or change direction in the plastic block. We’ve specifically been told that
the refracted ray in the block travels at an angle of 33 degrees to the line normal
to the surface. And so if we imagine that this is
the ray of light now traveling through the plastic block, then we can say that the
angle between that ray of light and the normal line, which we will call here 𝜃 sub
two, is equal to 33 degrees as we’ve been told in the question.
Now that we have drawn out the
problem, we need to work out the refractive index of the plastic that results in
this specific refraction. To do this, we need to recall
Snell’s law. This is given by the equation 𝑛
sub 𝑖 multiplied by sin 𝜃 sub 𝑖 equals 𝑛 sub 𝑟 multiplied by sin 𝜃 sub 𝑟,
where 𝑛 sub 𝑖 is the refractive index of the light ray’s initial material. 𝑛 sub 𝑟 is the refractive index
of the light ray’s final material. 𝜃 sub 𝑖 is the angle of
incident. And 𝜃 sub 𝑟 is the angle of
refraction.
We want to calculate the refractive
index of the plastic, which is the light ray’s final material. So we need to rearrange this
equation to make 𝑛 sub 𝑟 the subject. We can do this by dividing both
sides of the equation by sin 𝜃 sub 𝑟, which will leave us with 𝑛 sub 𝑟 equals 𝑛
sub 𝑖 sin 𝜃 sub 𝑖 over sin 𝜃 sub 𝑟.
Now, we are told in the question
that the light ray’s initial material is air. So 𝑛 sub 𝑖 is the refractive
index of air. We can recall that the refractive
index of air is very close to one and can be approximated as equal to exactly
one. So 𝑛 sub 𝑖 equals one. We also know that the angle of
incidence is this angle here. So 𝜃 sub 𝑖 is equal to 45
degrees. And the angle of refraction is this
angle here. So 𝜃 sub 𝑟 equals 33 degrees. Substituting these values into this
equation and grabbing a calculator, we find that 𝑛 sub 𝑟 is equal to 1.298 and so
on.
We need to give this answer to one
decimal place. So, if we round this up, we find
that 𝑛 sub 𝑟 is equal to 1.3. And we have arrived at the final
answer. The refractive index of the plastic
is 1.3.
