Snell's Window fisherman problem TIRF

[muddyjch]

Homework Statement

For a swimmer or fish, it appears that light from above the surface enters only through a circle, defined by an angle , theta as shown below:(See Diagram)
For the following, assume that the refractive index of water is 1.33 and that of air is 1.0.
(a) Suppose that the fish in the drawing above is 2m below the surface of a lake and is
looking upward. What is the angle defining the apparent window? (You must show
how to calculate this angle, not just look it up!)
(b) What is the diameter of the window the fish sees?
(c) Suppose that the fish is 6m from the bank of the lake, and there is a fisherman standing on the bank. Will the fish be able to see the fisherman? Will the fisherman be able to see the fish? Explain your answer with appropriate drawings and calculations.

Homework Equations

$$\theta$$_c=arcsin(n₁/n₂)

The Attempt at a Solution

For part a I got arcsin(1.00/1.33)=48.75 degrees. If you multiply by 2 that gives you $$\theta$$=97.5 degrees.
Part b tan 48.75=x/2m
x=2.28m
Part c i don't understand?

 

[anathema]

For part c, we need to consider the refraction of light at the air-water interface. When light travels from air to water, it bends towards the normal (the imaginary line perpendicular to the surface of the water). This means that the fisherman standing on the bank will appear to be at a shallower angle from the fish's perspective.

To calculate the angle at which the fisherman appears to the fish, we can use Snell's law:

n1sin\theta1 = n2sin\theta2

Where n1 is the refractive index of air (1.0), n2 is the refractive index of water (1.33), \theta1 is the angle of incidence (the angle at which light from the fisherman enters the water), and \theta2 is the angle of refraction (the angle at which light bends towards the normal inside the water).

Since we know that the fish is 6m from the bank, we can use basic trigonometry to calculate the angle of incidence:

tan\theta1 = 6m/x

Where x is the distance from the fisherman to the point where the light enters the water. Solving for x, we get x = 6m/tan\theta1.

Substituting this into Snell's law, we get:

n1sin\theta1 = n2sin\theta2

1.0 * sin(\theta1) = 1.33 * sin(\theta2)

Solving for \theta2, we get:

\theta2 = arcsin(1.0/1.33 * sin(\theta1))

Plugging in the value we calculated for \theta1, we get:

\theta2 = arcsin(1.0/1.33 * tan^-1(6m/x))

Now, to determine if the fish can see the fisherman and vice versa, we need to compare the angles at which they see each other. If the angle at which the fish sees the fisherman is smaller than the angle at which the fisherman sees the fish, then the fish will be able to see the fisherman but the fisherman will not be able to see the fish. If the opposite is true, then the fisherman will be able to see the fish but the fish will not be able to see the fisherman.

So, we need to compare the angle we calculated for \theta2 to the angle we calculated for \