Geometric Optics: Find Fish Apparent Position & Length in Fishtank

AI Thread Summary
A fish in a spherical glass fishtank is analyzed for its apparent position and length using geometric optics principles. The fish is 2cm long, located in water with an index of refraction of 1.33, while the glass has a thickness of 0.8cm and an index of 1.56. Initial calculations suggest the apparent length of the fish is 2.42cm, but there is confusion regarding the interpretation of results and the role of the glass. Participants discuss the need to consider both boundaries between water and glass, and glass and air, suggesting that the glass can be treated as a lens. The conversation emphasizes the importance of drawing diagrams and understanding the behavior of lenses, even if not yet covered in class.
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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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