Verifying Refraction of Coin in Water & Benzene

[KBriggs]

Hey,
this is just an exercise in my textbook. I think I have the solution, but I would just like for someone to verify my answer.

A coin rests on the bottom of a tank of water 1m deep, with a 20cm layer of benzene on top of the water. n for water is 1.33, for benzene it is 1.5. Find the apparent depth of the coin views from normal incidence in air above the benzene.

What I did was I split it up - I assumed an observer in the benzene, and proved that the ratio of apparent depth to real depth was:

$$\frac{d'}{d} = \frac{n_b}{n_w}=>d' = 1.128m$$ for an observer in benzene at the b-w interface looking down through water.

Then I treated the problem as though the whole tank was benzene and the object was at a depth 0f 0.2 + 1.128m. The analysis was the same, with the result of d'' = 0.885m.

Can someone verify these numbers for me?

Thanks :)

 

[Doc Al]

Your method and answer look good to me.

 

[kerimek]

Dear student, your solution seems to be correct. The apparent depth of the coin viewed from normal incidence in air above the benzene can be calculated by considering the refraction at the interface between air and benzene, and then at the interface between benzene and water. Your approach of splitting the problem into two parts is a valid method. Your calculated values of 1.128m and 0.885m for the apparent depth of the coin are in line with the expected values based on the refractive indices of water and benzene. Good job on your solution!