Solution

Statement

$13.3.1.$ The figures show objects and lens foci to a certain scale. Construct the images of these objects. What is the magnification in each case?

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 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 $f$ and $2f$, specifically at $d = \frac{5}{3}f$.

$$\Gamma = \frac{f}{d - f} = \frac{f}{\frac{5}{3}f - f} = \frac{f}{\frac{2}{3}f} = \frac{3}{2}$$

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

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

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

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