- #1
KBriggs
- 29
- 0
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 :)
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 :)
