91Ó°ÊÓ

• 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,$d=R/2$

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,

$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_1-n_2}{r}$ …(¾±)

The lateral magnification of the lens,

$m=\frac{h\text{'}}{h}$ …(¾±¾±)

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:

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

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 $1.14$.