A biconvex lens has radii of curvature, 20 cm20\text{ cm}20 cm each. If the refractive index of the material of the lens is 1.51.51.5, the power of the lens is
+2 D+2\text{ D}+2 D
+20 D+20\text{ D}+20 D
+5 D+5\text{ D}+5 D
Infinity
Power of lens is given by P=1f(m)P = \frac{1}{f(m)}P=f(m)1. Using the lens maker's formula: 1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)f1=(μ−1)(R11−R21). For a biconvex lens, R1=+20 cmR_1 = +20\text{ cm}R1=+20 cm and R2=−20 cmR_2 = -20\text{ cm}R2=−20 cm. Thus, 1f=(1.5−1)(120−1−20)=0.5(220)=0.5×110=120\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right) = 0.5 \left( \frac{2}{20} \right) = 0.5 \times \frac{1}{10} = \frac{1}{20}f1=(1.5−1)(201−−201)=0.5(202)=0.5×101=201. So, f=20 cm=0.2 mf = 20\text{ cm} = 0.2\text{ m}f=20 cm=0.2 m. Therefore, P=10.2=5 DP = \frac{1}{0.2} = 5\text{ D}P=0.21=5 D.
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