### Video Transcript

A light ray traveling in water of
refractive index 1.3 is incident on the flat surface of a plastic block with a
refractive index of 1.7, hitting the surface at an angle of 45 degrees from the line
normal to the surface. At what angle from the line normal
to the surface does the refracted ray in the block travel? Answer to the nearest degree.

This is a question about light
refracting as it travels from one medium to another. So let’s recall that Snell’s law
describes how light behaves as it passes between different media. Snell’s law states that for a light
ray passing from a medium with refractive index 𝑛 one to a medium with refractive
index 𝑛 two, the angle of incidence 𝜃 i is related to the angle of refraction 𝜃 r
by the formula 𝑛 one times sin 𝜃 i equals 𝑛 two times sin 𝜃 r.

To keep things organized, it’ll be
helpful to draw a diagram using the information provided by the question. Here in this sketch, we’ve shown in
orange the light ray traveling downward from the water to the plastic. We’ve labeled the respective
indices of refraction. Water’s index of refraction, 1.3,
is 𝑛 one, and the index of refraction of the plastic, 1.7, is 𝑛 two. We have also labeled the angle of
incidence, 𝜃 i, which equals 45 degrees. Remember that as we measure angles
here, they need to be measured from the line normal to the surface. Normal to the surface just means
perpendicular to the surface, and so this dashed line represents the normal
line.

Now, the question is asking us for
the angle of refraction, 𝜃 r. So we need to take Snell’s law and
rearrange it to make 𝜃 r the subject. To do this, let’s first divide both
sides by 𝑛 two, so that term cancels out of the right-hand side. Then, we can take the inverse sine
of both sides to undo the sine function on the right-hand side, thus isolating 𝜃
r. The equation we’ve ended up with
says that 𝜃 r equals the inverse sin of 𝑛 one times sin 𝜃 i divided by 𝑛
two. Since we already know the values of
all three of the variables on the right-hand side of this equation, we’re ready to
substitute them in and calculate our final answer. Let’s now clear some space on
screen to do this.

Subbing in the values of 𝑛 one, 𝑛
two, and 𝜃 i, we have that 𝜃 r equals the inverse sin of 1.3 times the sin of 45
degrees divided by 1.7. We can evaluate this by typing the
expression into a calculator, giving a result for 𝜃 r of 32.7333 et cetera
degrees.

Now, all that’s left to do is to
round this to the nearest degree. We find that to the nearest degree,
the angle of refraction 𝜃 r equals 33 degrees. This is our final answer to the
question.

If we want, we can then complete
our diagram by drawing in this refracted light ray, making sure that this angle 𝜃 r
is measured relative to the surface normal.