### Video Transcript

Fill in the blank. Light travels a distance quantified
by three π₯ meters in five π‘ seconds in air. If light travels a distance of π₯
meters in π‘ seconds in another medium, then the critical angle of this medium is
blank degrees. (A) 53.13, (B) 30.96, (C) 36.87,
(D) 59.04.

This question is asking us to
calculate the critical angle for some unknown medium. Letβs start by reminding ourselves
about the critical angle.

We can imagine a ray of light
traveling through a medium and then reaching the boundary of a second medium. When this happens, the light ray is
often refracted so that it passes through the second medium like this. However, if a light ray reaches the
new material at a particular angle of incidence, called the critical angle, which
weβll label π sub c, the light ray wonβt be transmitted through the second
medium. Instead, itβll travel along the
boundary, like this. This happens when the angle of
refraction is equal to 90 degrees.

To figure out the critical angle,
we need to recall Snellβs law of refraction. π sub i times the sin of π sub i
equals π sub r times the sin π sub r, where π sub i is the refractive index of
the first medium. π sub r is the refractive index of
the second medium. And π sub i and π sub r are the
angles of incidence and refraction.

Letβs replace the angle of
incidence with the critical angle π sub c and substitute in π sub r equals 90
degrees. This leaves us with π sub i times
the sin of π sub c equals π sub r times the sin of 90 degrees. The sin of 90 degrees equals
one. So this just becomes π sub i times
the sin of π sub c equals π sub r. Since weβre interested in the
critical angle, we can rearrange this to make the sin of π sub c the subject. sin of π sub c equals π sub r
over π sub i. So the sine of the critical angle
is equal to the refractive index of the second medium divided by the refractive
index of the first medium.

Recall that the refractive index of
a medium is equal to the speed of light in vacuum π divided by the speed of light
in the medium, which weβll call π£. Letβs find an expression for the
refractive index of the first medium π sub i. In this case, weβre told that the
medium is air. We know that light travels a
distance of three π₯ meters in a time of five π‘ seconds when in this medium. Using the formula speed equals
distance over time, we can find that the speed of light π£ in this medium is equal
to three π₯ over five π‘ meters per second. So the refractive index π sub i is
equal to π divided by three π₯ over five π‘. We can rewrite this slightly more
simply as π sub i equals five π times π‘ over three π₯.

Now letβs find an expression for
the refractive index of the second medium, π sub r. We know that light travels a
distance of π₯ meters in π‘ seconds in this medium. So, the speed of the light in this
medium is equal to π₯ over π‘ meters per second. To find the refractive index, we
simply divide this into the speed of light in a vacuum to get π sub r equals π
divided by π₯ over π‘ or π times π‘ over π₯.

Next, we can substitute these
expressions into our formula for the sine of the critical angle: sin of π sub c
equals π sub r over π sub i. Doing this, we find that the sine
of the critical angle is equal to π times π‘ over π₯ divided by five π times π‘
over three π₯. We can see that the factors of π
times π‘ over π₯ cancel out, leaving us with a value of three divided by five. This is equal to the sine of the
critical angle.

To find the critical angle itself,
we need to take the arcsine of both sides of this equation. This leaves us with the expression
π sub c equals the arcsin of three divided by five. Running this through our
calculator, we can find that the arcsin of three-fifths is equal to 36.8698 and so
on degrees. When rounded to two decimal places,
this gives us the answer 36.87 degrees. This corresponds to option (C). So the correct answer to this
question is (C). The critical angle of this medium
is 36.87 degrees.