Goldfish bowl index of refraction

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SUMMARY

The index of refraction of the aquarium walls can be determined using the magnification formula and the lens maker's equation. The magnification (m) is calculated as 0.72, derived from the thickness of the walls (4.30 mm) and the perceived thickness (3.10 mm). The equation used is n_1/s + n_2/s' = (n_2 - n_1)/R, where n_1 is the index of refraction of water (1.33). The user seeks clarification on the values of R, s, and s' to solve for the index of refraction (n_2) of the aquarium walls.

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  • Understanding of optics, specifically refraction and magnification
  • Familiarity with the lens maker's equation
  • Knowledge of the concept of object distance (s) and image distance (s')
  • Basic algebra skills for solving equations
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  • Learn about the concept of object and image distances in optics
  • Research how to calculate the radius of curvature (R) for different lens shapes
  • Explore practical examples of index of refraction calculations in various materials
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Homework Statement


To a fish in an aquarium, the 4.30 mm-thick walls appear to be only 3.10 mm thick.
What is the index of refraction of the walls?

Homework Equations


m=-s'/s
n_1/s + n_2 / s' = (n_2-n_1)/R
m: magnification
s: object distance
s': image distance


The Attempt at a Solution


m=(3.1/4.3)=.72
m=-s'/s
n_1/s + n_2 / s' = (n_2-n_1)/R
1.33/s + n_2/s' = (n_2-1.33)/R

I'm not sure what R, s, or s' would be.
If I plug in the numbers I received from determing the magnification, I receive this:

1.33/4.33 + n_2/3.1 = (n_2 - 1/33)/R
 
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I'm not sure if this is the correct equation to solve for the index of refraction or if I'm missing something. Any help is appreciated.
 

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