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+### Statement
+
+$7.2.4.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+Derive the formula for a long-focus thin lens:
+$$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$
+where $f$ is the focal length, $a$ is the distance from the electron source to the lens, and $b$ is the distance from the point where the electrons are focused to the lens.
+
+---
+
+### Solution
+
+Let us establish the geometric relations for the electron beam. We assume the paraxial approximation, meaning the angles are small enough that $\tan \theta \approx \theta$.
+
+The beam leaves a point source on the principal axis at a distance $a$ before the lens and strikes the lens at a vertical height $y$. The angle of the incident beam relative to the principal axis is:
+$$ \theta = \frac{y}{a} $$
+
+According to the problem description, the lens changes the beam's angle by an amount proportional to its deviation from the axis:
+$$ \Delta\theta = -\frac{y}{f} $$
+
+The new angle of the beam immediately after passing through the lens is:
+$$ \theta' = \theta + \Delta\theta = \theta - \frac{y}{f} $$
+
+After the lens, the beam converges to a point on the principal axis at a distance $b$. Since the beam is traveling downwards from a height $y$ to $0$ over a horizontal distance $b$, its angle relative to the axis is strictly negative:
+$$ \theta' = -\frac{y}{b} $$
+
+Equating the two expressions for $\theta'$:
+$$ -\frac{y}{b} = \frac{y}{a} - \frac{y}{f} $$
+
+Since the beam height $y \neq 0$, we can divide the entire equation by $y$:
+$$ -\frac{1}{b} = \frac{1}{a} - \frac{1}{f} $$
+
+Rearranging the terms yields the classical thin lens equation:
+$$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$
+
+#### Answer
+
+[Insert a concise answer or boxed result]