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 $13.1.14$
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:
$$fl_1 + fl_2 + 2f^2 = l_1l_2 + l_1f + l_2f + f^2$$$$2f^2 = l_1l_2 + f^2$$$$f^2 = l_1l_2$$
