Новое решение
en/13.3.9.md
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| + | ### Statement | ||
| + | |||
| + | $13.3.9.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | Let's define "magnification" as\ | ||
| + | $k = \frac{s'}{s}$ (1)\ | ||
| + | where $s'$ is the image distance from len and $s$ is the object distance from camera. | ||
| + | The rate of change for the object-to-len's distance is\ | ||
| + | $\dot{s} = -v$ (2)\ | ||
| + | it has negative sign due to the object goes towards the camera.\ | ||
| + | The relation with focal distance is given by\ | ||
| + | $\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$ (3)\ | ||
| + | Differentiating (3) respect to time,\ | ||
| + | $\frac{\dot{s}}{s^2}+\frac{\dot{s'}}{s'^2} = \frac{\dot{f}}{f^2}$ (4)\ | ||
| + | From (1),\ | ||
| + | $\dot{s'} = k\dot{s}$ (5)\ | ||
| + | Putting (2) and (5) into (4):\ | ||
| + | $\frac{\dot{f}}{f^2} = -\frac{v}{s^2}\left(1+\frac{1}{k}\right)$ (6)\ | ||
| + | From (3), and using (1),\ | ||
| + | $\frac{1}{f} = \frac{1}{s}\left(1+\frac{1}{k}\right)$ (7)\ | ||
| + | Substituting (7) into (6),\ | ||
| + | $\dot{f} = \frac{df}{dt} = \frac{-vk}{k+1}$ | ||
| + | the minus sign means that camera must reduce the focal distance at that rate. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $13.3.9.$ [Insert the problem statement] | |||
| ### Solution | |||
| Let's define "magnification" as\ | |||
| $k = \frac{s'}{s}$ (1)\ | |||
| where $s'$ is the image distance from len and $s$ is the object distance from camera. | |||
| The rate of change for the object-to-len's distance is\ | |||
| $\dot{s} = -v$ (2)\ | |||
| it has negative sign due to the object goes towards the camera.\ | |||
| The relation with focal distance is given by\ | |||
| $\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$ (3)\ | |||
| Differentiating (3) respect to time,\ | |||
| $\frac{\dot{s}}{s^2}+\frac{\dot{s'}}{s'^2} = \frac{\dot{f}}{f^2}$ (4)\ | |||
| From (1),\ | |||
| $\dot{s'} = k\dot{s}$ (5)\ | |||
| Putting (2) and (5) into (4):\ | |||
| $\frac{\dot{f}}{f^2} = -\frac{v}{s^2}\left(1+\frac{1}{k}\right)$ (6)\ | |||
| From (3), and using (1),\ | |||
| $\frac{1}{f} = \frac{1}{s}\left(1+\frac{1}{k}\right)$ (7)\ | |||
| Substituting (7) into (6),\ | |||
| $\dot{f} = \frac{df}{dt} = \frac{-vk}{k+1}$ | |||
| the minus sign means that camera must reduce the focal distance at that rate. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||