The radii of curvature of a lens are +20 cm and +30 cm . The...

Using the lens maker's formula:

\dfrac{1}{f} = \left(\dfrac{\mu_l}{\mu_m} -1 \right) \left(\dfrac{1}{R_1} - \dfrac{1}{R_2} \right)

Given $\mu_l = 1.6$, $\mu_m = 1.33 \approx$\dfrac{4}{3}, $R_1 = +20 \text{ cm}$, and $R_2 = +30 \text{ cm}$.

Substituting the values:

\dfrac{1}{f} = \left( \dfrac{1.6}{4/3} - 1 \right) \left( \dfrac{1}{20} - \dfrac{1}{30} \right)$<br><br>$\dfrac{1}{f} = \left( $\dfrac{4.8}{4} -$1 \right) \left( \dfrac{3 - 2}{60} \right)<br><br>\dfrac{1}{f} = (1.2 - 1) \times $\dfrac{1}{60}<br><br>\dfrac{1}{f} =$0.2 \times $\dfrac{1}{60} = \dfrac{1}{5} \times \dfrac{1}{60}<br><br>\dfrac{1}{f} = \dfrac{1}{300}<br><br>$f = 300 \text{ cm}