Video Transcript
A light ray traveling inside a tank
filled with a liquid of refractive index two times the square root of three falls on
the separating layer between the liquid and the air as shown in the figure
below. Which of the green lines shown will
be the path taken by the ray? Is it (A), (B), (C), or (D)?
In this question, we have a ray
that’s incident on an optical boundary. We need to determine the path of
the ray after it reaches the boundary and match it to one of the four lines labeled
A through D. Line A shows the ray transmitting
or passing through the boundary and changing direction significantly. Line B shows the ray transmitting
and changing direction slightly. Line C shows the ray skimming along
the boundary, neither transmitting nor reflecting. And line D shows the ray reflecting
at the boundary, staying inside the tank of liquid.
Recall that when a ray of light
entirely reflects, rather than transmits, at a boundary, it’s said to be
experiencing total internal reflection. Total internal reflection can only
occur if a ray is incident in a medium with a greater index of refraction than the
medium on the other side of the boundary. So let’s compare the refractive
indices of the two media in this situation.
We were told that the liquid in the
tank has a refractive index of two root three. So we’ll say that 𝑛 one equals two
root three, where 𝑛 denotes a refractive index. And the subscript one refers to
this first medium, the liquid in the tank. We need to compare this value to
the refractive index of the medium on the other side of the boundary, which we’ll
call 𝑛 two. Regarding this medium, we were told
that it is air. So let’s recall that air has an
index of refraction equal to one. This is the refractive index 𝑛
two. We already noted that in order for
total internal reflection to occur, 𝑛 one must be greater than 𝑛 two. In this question, 𝑛 one is indeed
greater than 𝑛 two, so total internal reflection is possible.
In order to determine whether this
is the fate of the ray in question, though, we need to calculate the critical angle
for this situation and compare it to the ray’s angle of incidence. Recall that the critical angle, 𝜃
sub 𝑐, can be found using the formula the sin of 𝜃 sub 𝑐 equals 𝑛 two divided by
𝑛 one. We can solve for the critical angle
by taking the arcsine of both sides in order to undo the sine function on the
left-hand side. Thus, we have that the critical
angle equals the arcsine of 𝑛 two over 𝑛 one.
Since we already know 𝑛 one and 𝑛
two, we can go ahead and substitute them in to calculate the critical angle. Doing this, the equation becomes 𝜃
sub 𝑐 equals the arcsine of one over two root three. Plugging this into a calculator
gives a result of about 16.8 degrees, so this is the critical angle. Next, we need to compare the
critical angle to the angle of incidence of the ray in the diagram. This will help us predict the ray’s
behavior at the boundary. If the angle of incidence, which
we’ll call 𝜃 sub one, is less than the critical angle, the ray will transmit. If 𝜃 sub one equals the critical
angle, then the ray will skim along the medium surface. And if 𝜃 sub one is greater than
the critical angle, the ray will undergo total internal reflection.
So let’s determine the value of 𝜃
sub one. Now, the diagram shows us an angle
labeled as 20 degrees. But we have to be careful not to
mistake this for the angle of incidence. That’s because this angle here is
measured with respect to the surface. But the angles we’re dealing with
need to be measured with respect to the surface normal. To represent the surface normal,
let’s draw in a dashed line that’s normal, or at 90 degrees, to the surface.
Now, it’s easier to see the ray’s
actual angle of incidence, and it’s equal to 90 minus 20 or 70 degrees. So we know that 𝜃 sub one equals
70 degrees, which is much greater than the critical angle. This means that the ray will, in
fact, undergo total internal reflection.
Referring back to the answer
options, we see that line D is the only path that shows total internal reflection of
the ray. So this must be the correct
choice. Therefore, our final answer is
option (D). The ray will experience total
internal reflection and take the path shown by line D.
