The power of a biconvex lens is 12 D and the radius of curvature of e…

The correct answer is Option (1) → 1.6

Given:

Power of lens $P = 12 \, D$

Radius of curvature of each surface $R = 10 \, cm = 0.1 \, m$

Lensmaker's formula for thin lens:

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

For biconvex lens: $R_1 = R$, $R_2 = -R$

$\frac{1}{f} = (n - 1) \left( \frac{1}{0.1} - \frac{1}{-0.1} \right) = (n - 1) \left( \frac{1}{0.1} + \frac{1}{0.1} \right) = (n - 1) \cdot 20$

Power $P = \frac{100}{f(\text{cm})} = \frac{100}{f(\text{cm})} \Rightarrow f = \frac{100}{12} \approx 8.33 \, cm = 0.0833 \, m$

Alternatively, using SI units: $P = \frac{100}{f(\text{cm})} \Rightarrow f = \frac{100}{12} \approx 8.33 \, cm = 0.0833 \, m$

From lensmaker's formula: $\frac{1}{f} = 20 (n - 1)$

$\frac{1}{0.0833} = 20 (n - 1)$

$12 = 20 (n - 1)$

$n - 1 = \frac{12}{20} = 0.6$

$n = 1.6$

Answer: Refractive index $n = 1.6$