Critical angle: find the depth of the fish looking up

In summary, the fish can see anything that is at the level of the surface or higher. If it is from above the water's surface, then would it just refract critically at the surface of the water and not reach the fish's eyes?
  • #1
Seneka
41
0

Homework Statement


A fish floats in water with its eye at the centre of an opaque walled full tank of water of circular cross section. When the fish look upwards, it can see a fish-eye view of the surrounding scene i.e. it is able to view the hemisphere of the scene above the water surface, and centred at the top of the tank. The diameter of the tank is 30 cm, and the critical angle for water is 48 degrees. At what depth below the surface of the water, d, must the fish be floating?
A) 16.7cm
B) 13.5cm
C) 11.2cm
D) 10.0cm

Homework Equations

The Attempt at a Solution


From the question you can tell that okay, I'm going to have to use the critical angle and since the fish's eyes are in the center we will get a right angle triangle with a side of 15 cm and an angle with 48 degrees from there I can use trigonometry to work out the side. Since I want the vertical I would have ended up with two answers
10.0cm and 13.5cm. The actual answer is 13.5cm.
Now I don't understand the questions. The mark scheme says "From the fishes eye to the rim of the tank makes an angle of 48o with the vertical. The emerging ray is horizontal (which is the requirement to see the horizon). Hence the depth of the fish is, d x tan 48o = 15.0, giving d."
"it is able to view the hemisphere of the scene above the water surface, and centered at the top of the tank." What does this mean? what does the hemisphere of the scene mean and centers at the top of the tank? If it is from above the water's surface, then would it just refract critically at the surface of the water and not reach the fish's eyes?
 

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  • #2
Seneka said:
"it is able to view the hemisphere of the scene above the water surface, and centered at the top of the tank." What does this mean?

Think of the "bowl of the sky", what the ancients thought was an actual sphere around us. The fish can see anything that is at the level of the surface or higher.

Seneka said:
If it is from above the water's surface, then would it just refract critically at the surface of the water and not reach the fish's eyes?
If it refracts, that means it penetrates the water and therefore could reach something in the water like the fish. What happens at the critical angle is that a ray going from water to air is refracted at 90 degrees, just along the surface. So in the other direction a ray along the surface will be refracted at the critical angle into the water.

The rays farthest from the normal, the rays with the largest angle, are the ones that go to the rim of the tank. So if the line from the fish to the surface at the rim is at the critical angle, he'll be able to see rays coming from the horizontal to that point. If he looks up at the surface at any other place inside the tank, the angle is less than the critical angle and the incident rays he sees will be at angles above the horizontal.

So the place that determines the limits of his vision is at the rim of the water, where the water meets the edge of the tank.

Does the explanation make sense to you now?
 
  • #3
Seneka said:
"it is able to view the hemisphere of the scene above the water surface, and centered at the top of the tank." What does this mean?
In my opinion, the question is garbage. The fish could be one inch below the center of the tank and see the same hemisphere as when it is deeper.
 
  • #4
Great! Your answer is making the question make more sense.
So is that a gap in my knowledge? Is it a fact that a refracted ray let's say ray 1 if there was to be an incident ray that is the same as ray 1 except in the opposite direction will produce a refracted ray that is the same as the incident ray to ray 1. It's indicated in my diagrams if that sentence was confusing.

RPinPA said:
If he looks up at the surface at any other place inside the tank, the angle is less than the critical angle and the incident rays he sees will be at angles above the horizontal.
So if the fish was above the 13.5 cm the angle would be greater so you would get total internal reflections and if he was below than it would refract out of the container but not at 90 degrees to the container.

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  • #5
jbriggs444 said:
In my opinion, the question is garbage. The fish could be one inch below the center of the tank and see the same hemisphere as when it is deeper.

Using trig isn't the only option 13.5 cm
 

1. What is critical angle?

Critical angle is the angle of incidence at which light passing through a medium is refracted to an angle of 90 degrees, resulting in total internal reflection.

2. How is critical angle related to finding the depth of a fish looking up?

When light enters water from air, it is refracted due to the change in density. The critical angle is the maximum angle at which light can enter water and still be refracted. By measuring the critical angle, we can calculate the depth of the water and therefore the depth of the fish looking up.

3. How is the critical angle calculated?

The critical angle can be calculated using the refractive index of the two media involved (in this case, air and water) using Snell's law: sin(critical angle) = refractive index of second medium / refractive index of first medium.

4. What factors can affect the critical angle?

The critical angle can be affected by the refractive index of the two media, the temperature of the water, and any impurities or particles in the water that may affect the light's path.

5. Can the critical angle method be used for finding the depth of any body of water?

Yes, the critical angle method is a reliable way to find the depth of any body of water as long as the mediums involved have known refractive indices and there are no external factors that may affect the light's path.

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