A student wants to know the height of a small bright slit and the focal length of a convex lens…

Step 1. The slit. $h = \sqrt{h_1h_2} = \sqrt{18\ \text{mm} \times 8.0\ \text{mm}} = 12.0\ \text{mm}$.

Step 2. The larger magnification. $m_1 = \dfrac{h_1}{h} = \dfrac{18\ \text{mm}}{12.0\ \text{mm}} = 1.50$. Check: $m_2 = \dfrac{8.0\ \text{mm}}{12.0\ \text{mm}} = 0.667$, and $1.50 \times 0.667 = 1$.

Step 3. The positions. $\dfrac{x_2}{x_1} = 1.50$ and $x_1 + x_2 = 120\ \text{cm}$. So $x_1 = \dfrac{120\ \text{cm}}{2.50} = 48.0\ \text{cm}$ and $x_2 = 72.0\ \text{cm}$. The lens moves $d = 72.0\ \text{cm} - 48.0\ \text{cm} = 24.0\ \text{cm}$.

Step 4. The focal length. $f = \dfrac{D^2 - d^2}{4D} = \dfrac{(120\ \text{cm})^2 - (24.0\ \text{cm})^2}{4 \times 120\ \text{cm}} = \dfrac{13\,824\ \text{cm}^2}{480\ \text{cm}} = 28.8\ \text{cm}$.