Question

A goldfish in a spherical fish bowl of radius R is at the level of the center C of the bowl and at distance $\mathit{R}\mathbf{/}\mathbf{2}$from the glass (Fig. 34-55). What magnification of the fish is produced by the water in the bowl for a viewer looking along a line that includes the fish and the center, with the fish on the near side of the center? The index of refraction of the water is $\mathbf{1}\mathbf{.}\mathbf{33}$. Neglect the glass wall of the bowl. Assume the viewer looks with one eye. (Hint: Equation 34-5 holds, but Eq. 34-6 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—that is, they deviate from that line by only small angles.)

Short answer

The magnification of the fish is $1.14$.

Step-by-step solution

The given data:

• The radius of the spherical bowl is $R$.
• Index of refraction of water,$n=1.33$
• The distance of the gold fish from the center,$d=R/2$

Understanding the concept of the magnification factor:

The ratio of the image size to the actual object size is called the lateral magnification of the lens. It can also be given as the negative ratio of the image distance to the object distance. Thus, the image distance can be determined by this concept.

Formulae:

The lens maker equation for a spherical surface is,

$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_1-n_2}{r}$ …(i)

The lateral magnification of the lens,

$m=\frac{h\text{'}}{h}$ …(ii)

Calculation of the image distance:

The object is at distance$R/2$fromthecenter. Use the specified formula where,$n{}_1=1.33$, $n_2=1.0$, and $p=R/2$.

If the surface faced by the object is convex, r is positive, and if it is concave, r is negative. Thus, using the given data in equation (i), the image distance can be given in terms of radius curvature can be given as follows:

$$\begin{gathered} \frac{1.33}{R/2}+\frac{1}{i}=\frac{1-1.33}{-R} \\ \frac{2\times 1.33}{R}+\frac{1}{i}=\frac{1-1.33}{-R} \\ \frac{2.66}{R}+\frac{1}{i}=\frac{0.33}{R} \\ \frac{2.33}{R}=\frac{-1}{i} \\ i=\frac{-R}{2.33} \\ \left|i\right|=\frac{R}{2.33} \end{gathered}$$

Considering the size of fish to be hand its virtual size to be $h\text{'}$, the magnification factor of the image due to the lens can be given using equation (ii) as:

$m=\frac{r-\left|i\right|}{p}$

Here, from the image you can say that

$h\text{'}=r-\left|i\right|$

Substitute known values in the above equation, and you have

$$\begin{gathered} m=\frac{R-\frac{R}{2.33}}{R/2} \\ =\frac{1-\frac{1}{2.33}}{1/2} \\ =\frac{1-0.43}{0.5} \\ =1.14 \end{gathered}$$

Hence, the value of the magnification factor is $1.14$.