Fish Bowl Optics: Find Power, Focus Length & Magnification

  • Thread starter Thread starter kpl
  • Start date Start date
kpl
Messages
6
Reaction score
0
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?
 

Attachments

Last edited:
Physics news on Phys.org
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.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
Back
Top