Geometric Optics: Find Fish Apparent Position & Length in Fishtank

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Homework Help Overview

The problem involves determining the apparent position and length of a fish in a spherical glass fishtank, considering the indices of refraction of water and glass. The fish is 2cm long, and the tank has a radius of 20cm with a thickness of 0.8cm.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss applying the lens formula and question the interpretation of results, particularly regarding the observer's position and the role of the glass thickness. There is also consideration of how to handle multiple boundaries between different media.

Discussion Status

Some participants have offered insights into the boundaries involved and suggested drawing diagrams to clarify the situation. Others express uncertainty about applying the lensmaker's equation and the implications of treating the glass as a lens.

Contextual Notes

There is mention of the complexity introduced by the glass as a shell and the potential for treating the water as a flat surface, which may not have been covered in the participants' studies yet.

whatisreality
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Homework Statement


A fish 2cm long is floating in a spherical glass fishtank with radius 20cm. The glass is 0.8cm thick and has index of refraction n=1.56. The index of refraction of water is 1.33. Find the apparent position and length of the fish.

Homework Equations

The Attempt at a Solution


I tried just applying the same equations I would normally:
##\frac{n_a}{s} + \frac{n_b}{s'} = \frac{n_b-n_a}{R}##
s is object distance, s' image distance, R radius of curvature. I used s=10cm, s' unknown, R=-10cm and got s' to be -156/11.
The magnification at this point is
##m=-\frac{n_as'}{n_bs}##
So m= 1.209... and the fish appears to be 2.42cm long.
Problem is, if this is the right approach I don't really know how to interpret the answers. Is the observer at the interface between materials? Where does 0.8cm come into it? Can I just apply the same equations again to find what the person outside the fishtank sees?
 
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I know it's more work, but it helps readers a lot if you post all your working. In the above, I can't tell what you are plugging in for the indices without trying to reproduce your answer.
 
haruspex said:
I know it's more work, but it helps readers a lot if you post all your working. In the above, I can't tell what you are plugging in for the indices without trying to reproduce your answer.
Well, I was only considering the first boundary between the water and the glass, so I used na=1.33 and nb=1.56.Then I don't know what to do about the second boundary between the glass and air.
 
whatisreality said:
Well, I was only considering the first boundary between the water and the glass, so I used na=1.33 and nb=1.56.Then I don't know what to do about the second boundary between the glass and air.
Since the glass is only a shell, it would probably more accurate to ignore the glass and just consider air and water.
To bring the glass into it, draw a diagram. You have the fish on one side, a curved piece of glass, viewer on the other side. What are the radii of curvature of the inside and outside of the glass? Regard it as a lens.
 
Does a lens of constant thickness behave according to the lensmakers equation? I think I can take the lens to be thin. We haven't actually covered lenses though...
 
whatisreality said:
Does a lens of constant thickness behave according to the lensmakers equation? I think I can take the lens to be thin. We haven't actually covered lenses though...
Yes, I don't see why the glass would not behave approximately as a lens. You know the radii.
But I hadn't noticed you have treated the water as a flat surface. That will behave as a lens too. Puzzled that you'd be given this question before covering lenses in your studies.
 
Oh. I didn't know that about the water. I think I'll wait until my professor goes through the answers! Thank you for helping :)
 

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