Geometrical optics. Photometry. Quantum nature of light > Rectilinear propagation and reflection of light > Problem 13.1.14

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

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

[$f^2 = l_1l_2$]