# Find magnification of a fish in a fish bowl

1. Oct 14, 2013

### MrMoose

1. The problem statement, all variables and given/known data

A goldfish in a spherical fish bowl of radius R is at the level of the center of the bowl and at distance R/2 from the glass. What magnification of the fish is produced by the water of the bowl for a viewer looking along a line that includes the fish and the center, from the fish's side of the center? The index of refraction of the water in the bowl is 1.33 Neglect the glass wall of the bowl. Assume the viewer looks with one eye. (Hint: Eq. 35-5 holds, but Eq. 35-6 does not. You need to work with a ray diagram of the situation and assume that the rays are close to the observer's line of sight.)

2. Relevant equations

m = h'/h (Eq 35-5)

m = -i/p (Eq 35-6)

n1/p + n2/i = (n2-n1)/r

n2*sin(θ2) = n1*sin(θ1)

Small angle assumption: Tanθ = Sinθ

3. The attempt at a solution

First solve for the image distance, i

n1/p + n2/i = (n2-n1)/r

i = n2/(n2/r - n1/r - n1/p)

The problem tells you that the object distance p = r/2

i = n2*r/(n2 - 3*n1)

Since n2 for air ~ 1

i = -r/(1 - 3*n1)

i is negative because the image is virtual

Second, find the relation between the object height, h, and the image height, h' (see attached drawing)

Tan(θ1) = h/p

Tan(θ2) = h'/i

The law of refraction states: n1*sin(θ1) = n2*sin(θ2)

Also, using the small angle assumption: Tanθ = Sinθ

n1*(h/p) = n2*(h'/i)

Solve for h'/h

h'/h = n1*i/(p*n2)

where n2~1, p = r/2 and i = -r/(1 - 3*n1)

h'/h = -2*n1/(1-3*n1) = 0.89

The correct answer is 1.14. Please help. Thanks in advance, MrMoose

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2. Oct 14, 2013

### TSny

Is there a sign convention for r in this equation?

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