The radii of curvature for a thin convex lens are 10 cm and 15 cm respectively. The focal length of the lens is 12 cm. The refractive index of the lens material is

The thin lens is placed in air, so we use the lens-maker’s formula

$$\frac{1}{f} \;=\; (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

where
$$f$$ = focal length of the lens,
$$n$$ = refractive index of lens material with respect to air,
$$R_1$$ = radius of curvature of the first surface (the surface that light meets first),
$$R_2$$ = radius of curvature of the second surface.

Sign convention (Cartesian):
• Distances measured to the right of the surface are positive.
• For a double-convex lens, the first surface bulges toward the left (center of curvature on the right), so $$R_1 \gt 0$$.
• The second surface bulges toward the right (center of curvature on the left), so $$R_2 \lt 0$$.

Given magnitudes:
$$|R_1| = 10 \text{ cm},\; |R_2| = 15 \text{ cm},\; f = 12 \text{ cm}$$

Applying the signs:
$$R_1 = +10 \text{ cm},\; R_2 = -15 \text{ cm}$$

Substitute into the lens-maker’s formula:

$$\frac{1}{12} = (n-1)\left(\frac{1}{10} - \frac{1}{-15}\right)$$

Simplify the bracket first:

$$\frac{1}{10} - \left(-\frac{1}{15}\right) = \frac{1}{10} + \frac{1}{15}$$

Take the LCM (30):

$$\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6}$$

Thus

$$\frac{1}{12} = (n-1)\left(\frac{1}{6}\right)$$

Cross-multiply:

$$(n-1) = \frac{1}{12} \times 6 = \frac{6}{12} = 0.5$$

Add 1 on both sides:

$$n = 1 + 0.5 = 1.5$$

Therefore, the refractive index of the lens material is $$1.5$$.

Option C is correct.