Reflection/Refraction Problem

  • Thread starter l46kok
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Homework Statement


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.


Homework Equations


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?
 

Answers and Replies

  • #2
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Can someone please.... help...
 
  • #3
andrevdh
Homework Helper
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Rays of light from a point on the fish (not a ray travelling vertically upwards) will be refracted away from the normal when it exits into the air. This means that a person who looks at these rays will conclude that they originated from a point higher up in the water than where they actually came from.

You need to determine the depth of this "apparent" point as viewed by such a person.
 

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