A tank whose bottom is a mirror is filled with water to a depth of 21.0cm. A small fish floats motionless 6.80cm under the surface of the water. Use index of refraction 1.33 for water and 1 for air.
What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence?
The Attempt at a Solution
First of all, I don't think I understand the question properly. Really, I think this is the most retardedly (Excuse me for the language) worded question ever.
I tried to solve it anyhow. Here's what I did :
Generally, for a spherical-relationship optic equation is
(n_a/s_a) + (n_b / s_b) = [(n_b) - (n_a)] / R
In this case, the fish tank's bottom surface is flat. So R (Radius of curvature) = infinity, which makes the second part 0.
So plugging in the values
(1.33 / 21.0 - 6.8) + (1 / s_b) = 0
and solving for s_b gives -10.67, but we cannot have negative distance, therefore, the answer is 10.67cm which is completely wrong.
can anyone help me what I did wrong in the steps?