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+### Statement
+
+$13.3.1.$ [Insert the problem statement]
+
+### 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
+
+![For problem $13.3.1$|123x456, 50%](https://sun9-42.userapi.com/s/v1/if2/0vMYHJAdUzU-5FUQuXusOc-XlOWZxHO5EMquAH29ZU8Ua2c82koQCMdGllJC-sFwdAttOn6hgYJtT6KHCgG44lZI.jpg?quality=95&as=32x18,48x27,72x41,108x62,160x91,240x137,360x205,480x274,540x308,573x327&from=bu&cs=573x0)
+
+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 $d = \frac{1}{3}|f|$.
+
+$$
+\Gamma = \frac{|f|}{\frac{1}{3}|f| + |f|} = \frac{|f|}{\frac{4}{3}|f|} = \frac{3}{4}
+$$
+
+#### Answer
+
+1. $\Gamma = 1/2$
+2. $\Gamma = 3/2$
+3. $\Gamma = 1/4$
+4. $\Gamma = 3/4$
+
+#### Answer
+
+[Insert a concise answer or boxed result]