Правка разделов «Statement», «Solution», «Answer»

JustABeast правка от
правка #17983 предыдущая #17982 ← раньше
@@ -1,13 +1,20 @@
### Statement
−$13.1.14.$ [Insert the problem statement]
+$13.1.14.$ Show that if the distances from the subject and image to the focus of a concave
+mirror are $l_1$ and $l_2$, then $$l_1 \cdot l_2 = f^2$$ where f is the focal length of the mirror.
### Solution
−
![For problem $13.1.14$ |1508x1495, 31%](../../img/13.1.14/scan-0.png)
+We are going to use the formula for a thin lens that relates the object distance $a$ image distance $b$ and focal length $f$. Look at the picture above as a referance: $$\frac{1}{f} = \frac{1}{a} + \frac{1}{b}$$
+We also know that $a = l_1 + f$ and $b = l_2 + f$. Putting this into our formula:
+$$\frac{1}{f} = \frac{1}{l_1 + f} + \frac{1}{l_2 + f}$$
+after simplifying:
+$$f = \frac{(l_1 + f)(l_2 + f)}{l_1 + l_2 + 2f}$$
+Multiplying both sides by $l_1 + l_2 + 2f$ gives us:
+$$fl_1 + fl_2 + 2f^2 = l_1l_2 + l_1f + l_2f + f^2$$
+$$2f^2 = l_1l_2 + f^2$$
+$$f^2 = l_1l_2$$
−
−
#### Answer
−[Insert a concise answer or boxed result]
+[$f^2 = l_1l_2$]