New solution

Luisito edited
View solution History
revision #18696 newer →
en/13.3.7.md +35 −0
@@ -0,0 +1,35 @@
+### Statement
+
+$13.3.7.$ [Insert the problem statement]
+
+### 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\
+$y_2'$ = 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)\
+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\
+$f = \frac{y_2's_2-y_1's_1}{y_2'-y_1'} = 3/7\;\rm{m}$
+
+#### Answer
+
+[Insert a concise answer or boxed result]