Edits to “Statement”, “Solution”, “Answer”
en/13.3.7.md
+3 −3
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $13.3.7.$ [Insert the problem statement] | ||
| + | $13.3.7.$ The image of an object on the frosted glass of the camera when photographing from a distance of 15 m reaches a height of 30 mm, and from a distance of 9 m - a height of 51 mm. Find the focal length of the lens. | ||
| ### Solution | |||
| Let's define some magnitudes:\ | |||
| $s_1$ = 15 m, first distance between object and camera\ | |||
| $y_1'$ = 30 mm, size of image on camera at first take\ | |||
| $s_2$ = 9 m, second distance between object and camera\ | |||
| @@ -11,7 +11,7 @@Solution | |||
| $y_2'$ = 51 mm, size of image on camera at second take\ | |||
| \ | |||
| It's known that\ | |||
| − | $\frac{1}{s_1}+\frac{}{s_1'}=\frac{1}{f}$ (1)\ | ||
| + | $\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{f}$ (1)\ | ||
| and the magnification is\ | |||
| $\frac{y_1'}{y}=\frac{s_1'}{s_1}$ (2)\ | |||
| and for second take, magnification is\ | |||
| $\frac{y_2'}{y}=\frac{s_2'}{s_2}$ (3)\ | |||
| and relation between distances also is given by:\ | |||
| $\frac{1}{s_2}+\frac{1}{s_2'}=\frac{1}{f}$ (4)\ | |||
| where $y$ is real size of the object and $s_i'$ ($i$ = 1,2) are the image-camera distance for each take.\ | |||
| Equaling (2) and (3):\ | |||
| $\frac{y_1's_1}{s_1'}=\frac{y_2's_2}{s_2'}$ (5)\ | |||
| Equaling (1) and (4):\ | |||
| $\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{s_2}+\frac{1}{s_2'}$ (6)\ | |||
| Separating $s_2'$ from (5):\ | |||
| $s_2' = \frac{y_2's_2s_1'}{y_1's_1}$ (7)\ | |||
| Putting (7) into (6):\ | |||
| $\frac{1}{s_1'} = \frac{y_2'(s_1-s_2)}{s_1(y_2's_2-y_1's_1)}$ (8)\ | |||
| Substituting (8) into (1), and developing\ | |||
| $f = \frac{y_2's_2-y_1's_1}{y_2'-y_1'} = 3/7\;\rm{m}$ | |||
| @@ -32,4 +32,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $f$ = 3/7 m | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $13.3.7.$ [Insert the problem statement] | $13.3.7.$ The image of an object on the frosted glass of the camera when photographing from a distance of 15 m reaches a height of 30 mm, and from a distance of 9 m - a height of 51 mm. Find the focal length of the lens. | ||
| ### Solution | ### Solution | ||
| Let's define some magnitudes:\ | Let's define some magnitudes:\ | ||
| $s_1$ = 15 m, first distance between object and camera\ | $s_1$ = 15 m, first distance between object and camera\ | ||
| $y_1'$ = 30 mm, size of image on camera at first take\ | $y_1'$ = 30 mm, size of image on camera at first take\ | ||
| $s_2$ = 9 m, second distance between object and camera\ | $s_2$ = 9 m, second distance between object and camera\ | ||
| @@ -11,7 +11,7 @@Solution | |||
| $y_2'$ = 51 mm, size of image on camera at second take\ | $y_2'$ = 51 mm, size of image on camera at second take\ | ||
| \ | \ | ||
| It's known that\ | It's known that\ | ||
| $\frac{1}{s_1}+\frac{}{s_1'}=\frac{1}{f}$ (1)\ | $\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{f}$ (1)\ | ||
| and the magnification is\ | and the magnification is\ | ||
| $\frac{y_1'}{y}=\frac{s_1'}{s_1}$ (2)\ | $\frac{y_1'}{y}=\frac{s_1'}{s_1}$ (2)\ | ||
| and for second take, magnification is\ | and for second take, magnification is\ | ||
| $\frac{y_2'}{y}=\frac{s_2'}{s_2}$ (3)\ | $\frac{y_2'}{y}=\frac{s_2'}{s_2}$ (3)\ | ||
| and relation between distances also is given by:\ | and relation between distances also is given by:\ | ||
| $\frac{1}{s_2}+\frac{1}{s_2'}=\frac{1}{f}$ (4)\ | $\frac{1}{s_2}+\frac{1}{s_2'}=\frac{1}{f}$ (4)\ | ||
| where $y$ is real size of the object and $s_i'$ ($i$ = 1,2) are the image-camera distance for each take.\ | where $y$ is real size of the object and $s_i'$ ($i$ = 1,2) are the image-camera distance for each take.\ | ||
| Equaling (2) and (3):\ | Equaling (2) and (3):\ | ||
| $\frac{y_1's_1}{s_1'}=\frac{y_2's_2}{s_2'}$ (5)\ | $\frac{y_1's_1}{s_1'}=\frac{y_2's_2}{s_2'}$ (5)\ | ||
| Equaling (1) and (4):\ | Equaling (1) and (4):\ | ||
| $\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{s_2}+\frac{1}{s_2'}$ (6)\ | $\frac{1}{s_1}+\frac{1}{s_1'}=\frac{1}{s_2}+\frac{1}{s_2'}$ (6)\ | ||
| Separating $s_2'$ from (5):\ | Separating $s_2'$ from (5):\ | ||
| $s_2' = \frac{y_2's_2s_1'}{y_1's_1}$ (7)\ | $s_2' = \frac{y_2's_2s_1'}{y_1's_1}$ (7)\ | ||
| Putting (7) into (6):\ | Putting (7) into (6):\ | ||
| $\frac{1}{s_1'} = \frac{y_2'(s_1-s_2)}{s_1(y_2's_2-y_1's_1)}$ (8)\ | $\frac{1}{s_1'} = \frac{y_2'(s_1-s_2)}{s_1(y_2's_2-y_1's_1)}$ (8)\ | ||
| Substituting (8) into (1), and developing\ | Substituting (8) into (1), and developing\ | ||
| $f = \frac{y_2's_2-y_1's_1}{y_2'-y_1'} = 3/7\;\rm{m}$ | $f = \frac{y_2's_2-y_1's_1}{y_2'-y_1'} = 3/7\;\rm{m}$ | ||
| @@ -32,4 +32,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $f$ = 3/7 m | ||