Solution

To analytically verify the magnification $\Gamma$ obtained from the geometrical construction, we can use the thin lens formula and the expression for magnification $\Gamma = \frac{f}{|d \pm f|}$, where $d$ is the distance from the object to the lens, and $f$ is the positive focal length. By determining the ratio of $d$ to $f$ directly from the scale of the given figures, we can calculate the exact values:

1. Figure (a): Converging lens. The distance to the object is $d = 3f$.

$$\Gamma = \frac{f}{d - f} = \frac{f}{3f - f} = \frac{1}{2}$$

2. Figure (b): Converging lens. The object is located between $0$ and $f$, specifically at $d = \frac{1}{3}f$.

$$\Gamma = \frac{f}{f-d} = \frac{f}{f-\frac{1}{3}f} = \frac{f}{\frac{2}{3}f} = \frac{3}{2}$$

3. Figure (c): Diverging lens. For a diverging lens, the magnification formula is $\Gamma = \frac{f}{d + f}$. The distance to the object is $d = 3f$.

$$\Gamma = \frac{f}{3f + f} = \frac{1}{4}$$

4. Figure (d): Diverging lens. The distance to the object is $d = \frac{1}{3}f$.

$$\Gamma = \frac{f}{\frac{1}{3}f + f} = \frac{f}{\frac{4}{3}f} = \frac{3}{4}$$