The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Its power will be :

2.5 D

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2 D

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1.5 D

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1 D

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Solution

The correct option is B 2.5 D
$P = \frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
Here, $μ = 1.5$ , $R_{1}$ = 40 cm = 0.4cm , $R_{2}$ = -0.4 cm
$P = (1.5 - 1) [\frac{1}{0.4} + \frac{1}{0.4}] = 2.5 D$

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