Lenses are fabricated from a material having a refractive index of...

Using the lens maker's formula:

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

Given $\mu = 1.50$, the term $(\mu - 1) = 0.5 =$\dfrac{1}{2}. The magnitudes of the radii of curvature are $20 \text{ cm}$ and $30 \text{ cm}$.

Depending on the type of lens (biconvex, biconcave, concavo-convex, or convexo-concave), the signs of $R_1$ and $R_2$ can be positive or negative. The possible values for the term \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right) are:<br><br>Case 1:R_1$and$R_2$have opposite signs.<br><br>\dfrac{1}{R_1} - \dfrac{1}{R_2} = \pm \left( \dfrac{1}{20} + \dfrac{1}{30} \right) = \pm \left( \dfrac{3 + 2}{60} \right) = \pm \dfrac{1}{12} \text{ cm}^{-1}<br><br>Substituting into the formula:<br><br>\dfrac{1}{f} = \dfrac{1}{2} \left( \pm \dfrac{1}{12} \right) = \pm \dfrac{1}{24} \text{ cm}^{-1} \Rightarrow$f = \pm 24 \text{ cm}$<br><br>Case 2:$R_1$and$R_2$have the same sign.<br><br>\dfrac{1}{R_1} - \dfrac{1}{R_2} = \pm \left( \dfrac{1}{20} - \dfrac{1}{30} \right) = \pm \left( \dfrac{3 - 2}{60} \right) = \pm \dfrac{1}{60} \text{ cm}^{-1}<br><br>Substituting into the formula:<br><br>\dfrac{1}{f} = \dfrac{1}{2} \left( \pm \dfrac{1}{60} \right) = \pm \dfrac{1}{120} \text{ cm}^{-1} \Rightarrow$f = \pm 120 \text{ cm}$<br><br>Therefore, the possible focal lengths are$\pm 24 \text{ cm}$and$\pm 120 \text{ cm}$.$