Edits to “Statement”, “Answer”
en/7.2.5.md
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| @@ -1,13 +1,8 @@ | |||
| ### Statement | |||
| − | $7.2.5.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| At what distance from the lens axis does an electron beam focus if it originates from point $A$ with coordinates $-x_0$ and $y_0$, and passes through a long-focus lens with focal length $f$ located at the origin? Assume $y_0 \ll x_0, f$. | |||
| + |  | ||
| ### Solution | |||
| Let us use the formula for a long-focus thin lens, as derived in [Problem 7.2.4](https://savchenkosolutions.com/en/7.2.4): | |||
| $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$ | |||
| In our given coordinate system, the electron source is located at $x = -x_0$, so the distance to the lens is $a = x_0$. Substituting this into the lens equation, we can find the distance $b$ (the coordinate along the $x$-axis where the electrons focus): | |||
| $$ \frac{1}{x_0} + \frac{1}{b} = \frac{1}{f} \implies \frac{1}{b} = \frac{1}{f} - \frac{1}{x_0} = \frac{x_0 - f}{f x_0} $$ | |||
| $$ b = \frac{f x_0}{x_0 - f} $$ | |||
| Electrons passing through the center of the lens (the origin) are not deflected. Using similar triangles for this central trajectory, we can relate the coordinates of the source and the focusing point: | |||
| $$ \frac{y}{b} = \frac{y_0}{-x_0} \implies y = -\frac{y_0 b}{x_0} $$ | |||
| Substituting the expression for $b$ into this equation yields: | |||
| $$ y = -\frac{y_0}{x_0} \left( \frac{f x_0}{x_0 - f} \right) $$ | |||
| $$ y = -\frac{y_0 f}{x_0 - f} $$ | |||
| *Note: The negative sign indicates that the focus point is located below the principal axis, which is consistent with the physical behavior of a converging lens focusing a beam when $x_0 > f$. The absolute distance from the axis is $\frac{y_0 f}{x_0 - f}$.* | |||
| @@ -30,7 +25,3 @@Solution | |||
| ### Answer | |||
| $$ y = -\frac{y_0 f}{x_0 - f} $$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,13 +1,8 @@ | |||
| ### Statement | ### Statement | ||
| $7.2.5.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| At what distance from the lens axis does an electron beam focus if it originates from point $A$ with coordinates $-x_0$ and $y_0$, and passes through a long-focus lens with focal length $f$ located at the origin? Assume $y_0 \ll x_0, f$. | At what distance from the lens axis does an electron beam focus if it originates from point $A$ with coordinates $-x_0$ and $y_0$, and passes through a long-focus lens with focal length $f$ located at the origin? Assume $y_0 \ll x_0, f$. | ||
|  | |||
| ### Solution | ### Solution | ||
| Let us use the formula for a long-focus thin lens, as derived in [Problem 7.2.4](https://savchenkosolutions.com/en/7.2.4): | Let us use the formula for a long-focus thin lens, as derived in [Problem 7.2.4](https://savchenkosolutions.com/en/7.2.4): | ||
| $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$ | $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{f} $$ | ||
| In our given coordinate system, the electron source is located at $x = -x_0$, so the distance to the lens is $a = x_0$. Substituting this into the lens equation, we can find the distance $b$ (the coordinate along the $x$-axis where the electrons focus): | In our given coordinate system, the electron source is located at $x = -x_0$, so the distance to the lens is $a = x_0$. Substituting this into the lens equation, we can find the distance $b$ (the coordinate along the $x$-axis where the electrons focus): | ||
| $$ \frac{1}{x_0} + \frac{1}{b} = \frac{1}{f} \implies \frac{1}{b} = \frac{1}{f} - \frac{1}{x_0} = \frac{x_0 - f}{f x_0} $$ | $$ \frac{1}{x_0} + \frac{1}{b} = \frac{1}{f} \implies \frac{1}{b} = \frac{1}{f} - \frac{1}{x_0} = \frac{x_0 - f}{f x_0} $$ | ||
| $$ b = \frac{f x_0}{x_0 - f} $$ | $$ b = \frac{f x_0}{x_0 - f} $$ | ||
| Electrons passing through the center of the lens (the origin) are not deflected. Using similar triangles for this central trajectory, we can relate the coordinates of the source and the focusing point: | Electrons passing through the center of the lens (the origin) are not deflected. Using similar triangles for this central trajectory, we can relate the coordinates of the source and the focusing point: | ||
| $$ \frac{y}{b} = \frac{y_0}{-x_0} \implies y = -\frac{y_0 b}{x_0} $$ | $$ \frac{y}{b} = \frac{y_0}{-x_0} \implies y = -\frac{y_0 b}{x_0} $$ | ||
| Substituting the expression for $b$ into this equation yields: | Substituting the expression for $b$ into this equation yields: | ||
| $$ y = -\frac{y_0}{x_0} \left( \frac{f x_0}{x_0 - f} \right) $$ | $$ y = -\frac{y_0}{x_0} \left( \frac{f x_0}{x_0 - f} \right) $$ | ||
| $$ y = -\frac{y_0 f}{x_0 - f} $$ | $$ y = -\frac{y_0 f}{x_0 - f} $$ | ||
| *Note: The negative sign indicates that the focus point is located below the principal axis, which is consistent with the physical behavior of a converging lens focusing a beam when $x_0 > f$. The absolute distance from the axis is $\frac{y_0 f}{x_0 - f}$.* | *Note: The negative sign indicates that the focus point is located below the principal axis, which is consistent with the physical behavior of a converging lens focusing a beam when $x_0 > f$. The absolute distance from the axis is $\frac{y_0 f}{x_0 - f}$.* | ||
| @@ -30,7 +25,3 @@Solution | |||
| ### Answer | ### Answer | ||
| $$ y = -\frac{y_0 f}{x_0 - f} $$ | $$ y = -\frac{y_0 f}{x_0 - f} $$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||