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Radius of curvature of one surface of double convex lens is three times of the other. If focal length of the lens is 30 cm and refractive index of the lens material is3/2, then radius of curvature are

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60 cm

80 cm

40 cm

20 cm

answer is A.

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Detailed Solution

$R_{1} = 3 R, R_{2} = - R, f = 30 cm, μ = \frac{3}{2}$

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{30} = (\frac{3}{2} - 1) (\frac{1}{3 R} + \frac{1}{R})$

$\frac{1}{30} = \frac{1}{2} \times (\frac{1 + 3}{3 R}) = \frac{1}{2} \times \frac{4}{3 R} \Rightarrow \frac{1}{30} = \frac{2}{3 R}$

$R = 20$

$Since R_{2} = 20 cm, R_{1} = 3 R = 3 (20) = 60 cm$

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