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