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Magnification of a fish in a spherical bowl due to water

  1. Dec 5, 2014 #1
    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.

    (Answer : 1.14)

    2. Relevant equations

    $$ m\ =\ \frac{h'}{h} $$
    where ##h'## is the height of the image , ##h## the height of the object.
    $$ \frac{\mu_2}{v}-\frac{\mu_1}{u}\ =\ \frac{\mu_1-\mu_2}{R} $$
    where all distances are from the center of the spherical medium following the sign convention such that distances to the left of origin are negative and to the right are positive.
    ##\mu_2## is the refractive index of air surrounding the fish bowl while ##\mu_1## is that of the water.
    $$ \mu_1 \sin{i} = \mu_2 \sin{r} $$

    3. The attempt at a solution


    $$ \begin{align}
    &\frac{\mu_2}{v}-\frac{\mu_1}{u}\ =\ \frac{\mu_1-\mu_2}{R} \\
    \text{Applying sign convention ,} \\
    &\frac{\mu_2}{-v}-\frac{\mu_1}{-u}\ =\ \frac{\mu_1-\mu_2}{-R} \\
    \implies &\frac{\mu_2}{v}-\frac{\mu_1}{u}\ =\ \frac{\mu_1-\mu_2}{R} \\
    \implies &\frac{\mu_2}{v}\ =\ \frac{\mu_1-\mu_2}{R}+\frac{\mu_1}{u} \\
    \text{Substituting the value of u ,} \\
    \implies &\frac{\mu_2}{v}\ =\ \frac{\mu_1-\mu_2}{R}+\frac{2 \mu_1}{R} \\
    \implies &\frac{\mu_2}{v}\ =\ \frac{3 \mu_1-\mu_2}{R} \\
    \implies &v\ =\ \frac{R}{3 \mu_1-\mu_2}
    \text{Now from the figure , } \\
    &\frac{h'}{h}\ =\ \frac{R-v}{R-u} \\
    \implies &\frac{h'}{h}\ =\ \frac{R-\frac{R}{3 \mu_1-\mu_2}}{R/2} \\
    \implies &m \ =\ \frac{h'}{h}\ =\ 2\frac{3\mu_1-2\mu_2}{3\mu_1-\mu_2} \\
    \text{Now , plugging in the values , we get ,} \\
    &m \ =\ 2\frac{3\times 1.33-2\times 1}{3\times 1.33 - 1} \\
    \end{align} $$
    So ,$$m = 1.33$$
    Which is not the same as the answer ##1.14##.

    So , where did I go wrong ? Is there another way of solving this ?

    P.S : I found that this same question was asked before , there the OP used a different method and the thread was not answered.
  2. jcsd
  3. Dec 6, 2014 #2


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    Check that step. (Btw, you lost a factor ##\mu_2## at one point, but since it is 1 it didn't matter.)
  4. Dec 6, 2014 #3
    Oops, I didn't noticed that , Also about that step , it is right if that factor of ##\mu_2## is included.
    Corrected steps. ( Is is not possible to edit the first post of a thread in the new forums ? I didn't find an "Edit" button. )
    $$ v = \frac{R \mu_2}{3\mu_1-\mu/2} $$
    $$ \begin{align}
    m &= \frac{R - v}{R-u} \\
    &= \frac{R-\frac{R \mu_2}{3\mu_1-\mu_2}}{R-\frac{R}{2} } \\
    &= \frac{R(3\mu_1-\mu_2)-R \mu_2}{R/2 (3\mu_1-\mu_2) } \\
    &= 2\frac{3\mu_1-2 \mu_2}{3\mu_1-\mu_2}
    \end{align} $$
  5. Dec 6, 2014 #4


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    Yes, you're right.
    I believe the problem is your very first equation. The derivation of that assumes the object is external to the lens. As I recall (from 50 years ago), one draws a diagram with one ray travelling from the tip of the object parallel to the axis, and a second passing through the centre of the lens. With an external object, the second ray is effectively straight because the two lens surfaces are parallel there. With an object inside the lens this fails.
    There's probably some standard treatment of this case, but I just did it from first principles and got the right answer.
    Some labels: O is centre of sphere, A is centre of goldfish, B is top of goldfish. Line OA meets glass at C, OB meets glass at D.
    Angle AOB is theta, so ACB is also theta (isosceles).
    Image forms at E (on line OC) to F (on OD). Angle OCF is phi. Index ratio is n = sin(phi)/sin(theta). Assume small angles, so =phi/theta.
    h = AB = OA tan theta, h' = EF.
    FC = h'/ sin phi.
    Sine rule: FC/sin theta = OC/sin(theta+phi).
    If you put all that together, you should get 2n/(n+1) as the answer.
  6. Dec 7, 2014 #5
    Thank you very much .
    I've been stuck with this problem for a while.

    > I believe the problem is your very first equation. The derivation of that assumes the object is external to the lens ...

    I think you mean the derivation of the lens formula not the formula concerning refraction just by a curved surface which is the formula in question. T
    he way we derived it is considering a point object on the axis , whose image is another point on the axis.
    The derivation involved two types of approximations :
    * Small angle approximation for sin and tan.
    * Assuming that the tip of the perpendicular to the axis drawn from the point of incidence is so close to the "center/pole" of the surface that they coincide.
    (Similar to
    Anyway , thanks and I guess I'll have to be careful applying the refraction formula.
    Last edited: Dec 7, 2014
  7. Dec 9, 2014 #6


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    This has been bothering me. Both approaches seem right, yet yield different answers.
    But I might have cracked it. I think there's a sign wrong here:
    ##\frac{\mu_2}{v}-\frac{\mu_1}{u}\ =\ \frac{\mu_1-\mu_2}{R}##
    I have ##\frac{\mu_1}{u}-\frac{\mu_2}{v} =\ \frac{\mu_1-\mu_2}{R}##.
    Comparing with the Pink Monkey set up is a bit tricky because in that example the object is in air, and v > R.
  8. Dec 17, 2014 #7
    Sorry for not replying , I am not a frequent visitor for physicsforums and I didn't looked after your approach worked for me.
    Yes , I think you are right , I rechecked the formula , I wrote the original formula wrong , I got the right answer now. :s:s
    That was the first result I got for a derivation but a similar derivation could be done for this specific case as well even if v > R.
    Anyway , Big Thanks :)
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