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