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| ### Statement |
| ### Statement |
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| $7.2.4.$ [Insert the problem statement] |
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| ### Solution |
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| ### Statement |
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| Derive the formula for a long-focus thin lens: |
| Derive the formula for a long-focus thin lens: |
| $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$ |
| $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$ |
| 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. |
| 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. |
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| ### Solution |
| ### Solution |
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| Let us establish the geometric relations for the electron beam. We assume the paraxial approximation, meaning the angles are small enough that $\tan \theta \approx \theta$. | | Let us establish the geometric relations for the electron beam. We assume the paraxial approximation, meaning the angles are small enough that $\tan \theta \approx \theta$. |
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| 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: | | 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: |
| $$ \theta = \frac{y}{a} $$ | | $$ \theta = \frac{y}{a} $$ |
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| According to the problem description, the lens changes the beam's angle by an amount proportional to its deviation from the axis: | | According to the problem description, the lens changes the beam's angle by an amount proportional to its deviation from the axis: |
| $$ \Delta\theta = -\frac{y}{f} $$ | | $$ \Delta\theta = -\frac{y}{f} $$ |
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| The new angle of the beam immediately after passing through the lens is: | | The new angle of the beam immediately after passing through the lens is: |
| $$ \theta' = \theta + \Delta\theta = \theta - \frac{y}{f} $$ | | $$ \theta' = \theta + \Delta\theta = \theta - \frac{y}{f} $$ |
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| 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: | | 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: |
| $$ \theta' = -\frac{y}{b} $$ | | $$ \theta' = -\frac{y}{b} $$ |
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| Equating the two expressions for $\theta'$: | | Equating the two expressions for $\theta'$: |
| $$ -\frac{y}{b} = \frac{y}{a} - \frac{y}{f} $$ | | $$ -\frac{y}{b} = \frac{y}{a} - \frac{y}{f} $$ |
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| Since the beam height $y \neq 0$, we can divide the entire equation by $y$: | | Since the beam height $y \neq 0$, we can divide the entire equation by $y$: |
| $$ -\frac{1}{b} = \frac{1}{a} - \frac{1}{f} $$ | | $$ -\frac{1}{b} = \frac{1}{a} - \frac{1}{f} $$ |
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| Rearranging the terms yields the classical thin lens equation: | | Rearranging the terms yields the classical thin lens equation: |
| $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$ | | $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$ |
