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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?