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?
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?