A biconvex lens has radii of curvature, $20 ext{ cm}$ each. If the refractive index of the material of the le

Question

A biconvex lens has radii of curvature, $20	ext{ cm}$ each. If the refractive index of the material of the lens is $1.5$, the power of the lens is

Accepted Answer

Power of lens is given by $P = \frac{1}{f(m)}$. Using the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$. For a biconvex lens, $R_1 = +20	ext{ cm}$ and $R_2 = -20	ext{ cm}$. Thus, $\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} ight) = 0.5 \left( \frac{2}{20} ight) = 0.5 	imes \frac{1}{10} = \frac{1}{20}$. So, $f = 20	ext{ cm} = 0.2	ext{ m}$. Therefore, $P = \frac{1}{0.2} = 5	ext{ D}$.

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