According to the Lens Maker's Formula for a convex lens in air: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$

Given: Focal length $f = +25 \text{ cm}$ (positive for convex lens) Refractive index $\mu = 1.5$

• Radii relationship: Let $R_1 = R$ and $R_2 = 2R$. Since it is a double convex lens, the signs of the radii relative to the incident light are opposite (Cartesian sign convention: $R_1$ is positive, $R_2$ is negative, but in the standard formula form for convex lens magnitudes add up: $\frac{1}{R_1} - \frac{1}{-R_2} = \frac{1}{R_1} + \frac{1}{R_2}$). Strictly applying signs: $\frac{1}{f} = (1.5 - 1)(\frac{1}{R} - \frac{1}{-2R})$.

Calculation: $\frac{1}{25} = (0.5) \left( \frac{1}{R} + \frac{1}{2R} \right)$ $\frac{1}{25} = \frac{1}{2} \left( \frac{2 + 1}{2R} \right)$ $\frac{1}{25} = \frac{1}{2} \cdot \frac{3}{2R} = \frac{3}{4R}$ $4R = 75 \implies R = \frac{75}{4} = 18.75 \text{ cm}$

The second radius is $2R = 2 \times 18.75 = 37.5 \text{ cm}$.

Thus, the radii are $18.75 \text{ cm}$ and $37.5 \text{ cm}$.