Edit to “Statement”
en/13.3.9.md
+1 −1
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $13.3.9.$ [Insert the problem statement] | ||
| + | $13.3.9.$ The subject is moving towards the movie camera at a speed of $v$. How fast do you need to change the focal length of the lens and the depth of the movie camera so that the image size remains the same if the magnification given by the movie camera is $k$? | ||
| ### 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 | |||
| $|\dot{f}| =\frac{vk}{k+1}$ | |||
| unchanged lines 23 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $13.3.9.$ [Insert the problem statement] | $13.3.9.$ The subject is moving towards the movie camera at a speed of $v$. How fast do you need to change the focal length of the lens and the depth of the movie camera so that the image size remains the same if the magnification given by the movie camera is $k$? | ||
| ### 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. | ||
| #### Answer | #### Answer | ||
| $|\dot{f}| =\frac{vk}{k+1}$ | $|\dot{f}| =\frac{vk}{k+1}$ | ||
| unchanged lines 23 | |||