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 [2022]

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Solution

Q.
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 [2022]

1 +2D

2 +20D

3 +5D

4 infinity

Ans.
(3)

Given, refractive index of lens = 1.5

For biconvex lens, $R_{1} = + R, R_{2} = - R$

$∴$ $R_{1} = + 20 cm and R_{2} = - 20 cm$

According to lens maker’s formula,

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$    $∴$ $\frac{1}{f} = (1.5 - 1) (\frac{1}{20} - \frac{1}{(- 20)}) = 0.5 (\frac{1}{20} + \frac{1}{20})$

or $\frac{1}{f} = \frac{1}{20} \Rightarrow f = 20 cm$

Power of lens (in dioptre),   $P = \frac{100}{focal length f (in cm)}$

$∴$    $P = \frac{100}{20} = + 5 D$

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