The radii of curvature of the two surfaces of a concave lens are 20 cm each. The refractive index of the material of the lens is 1.5. The power of the lens would be-

-5 D

The correct answer is Option (1) → -5 D

Given:

Radii of curvature = $R = 20\text{ cm} = 0.20\text{ m}$

Refractive index = $\mu = 1.5$

Lensmaker formula (power):

$P = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

For a symmetric biconcave lens with equal radii,

$P = -\frac{2(\mu-1)}{R}$

Substitute values:

$P = -\frac{2(1.5-1)}{0.20} = -\frac{2\times 0.5}{0.20} = -\frac{1}{0.20} = -5\ \text{D}$

The power of the lens is $-5\ \text{dioptre}$.