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:
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{f}$$