- #1

- 36

- 4

## Homework Statement

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)

## Homework 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.

$$ \mu_1 \sin{i} = \mu_2 \sin{r} $$

## 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.