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Fish in a Fish bowl?

  • Thread starter kpl
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kpl
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Q. There is a fish in a spherical fish bowl of radius r filled with water. The fish is at a distance l from the surface of the tank. An observer in air views the fish at a distance d from the tank.

Please see the attachment for the image.

I am supposed to find the optical power and focus length of the water-air interface, the magnification power of the fish tank and an expression for the distance (L) between the obsever and optical image of the fish.

Here are the attempts I have made:

Optical power = (n1-n2)/(-r)

Focus Length = f'=n2/P = -n2r/(n1-n2)
I used f' because I think it is a virtual image?

Magnification Power = -(Image Distance/Object Distance) =-((L-d)/-r)

L=(-n1/r + n2/l)

Can anyone tell me if these are correct or if I am at least going along in the right direction?
 

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Answers and Replies

Try the refraction at a single surface formula
n1/s+n2/s'=(n1-n2)/r

here, n1=1.33 (water); N2=1 (air)
and note the sign of s should be negative as the fish is not on the incident-light side of the surface.
 

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