Question: 3: A biconvex lens has radii of curvature, 20 cm each. If the refractive index of the material of the lens is 1.5 , the power of the lens is
(1) +5 \mathrm{D}
(2) Infinity
(3) +2 \mathrm{D}
(4) +20 D
Answer: Option (1)
Explanation:
For a thin lens, the focal length is given by the lens maker’s formula
\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right).
For a biconvex lens, the first surface has radius R_1 = +20 \text{ cm}
and the second surface has radius R_2 = -20 \text{ cm},
according to the sign convention.
The refractive index of the lens material is \mu = 1.5.
Substituting the values,
\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \left(-\frac{1}{20}\right)\right).
\frac{1}{f} = 0.5 \times \frac{2}{20} = \frac{1}{20}.
Thus, the focal length of the lens is f = 20 \text{ cm} = 0.20 \text{ m}.
The power of a lens is given by P = \frac{1}{f},
where f is in metres.
P = \frac{1}{0.20} = +5 \mathrm{D}.
Hence, the power of the biconvex lens is +5 \mathrm{D}.