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Reflection/Refraction Problem

  • Thread starter l46kok
  • Start date
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1. 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.


2. Homework Equations
What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence?


3. 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

296
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Can someone please.... help...
 
andrevdh
Homework Helper
2,126
116
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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