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When an object is placed at a distance of 20 cm from a equibiconvex lens a real image of it is formed at a distance 20 cm from the lens if the radius of curvature of each surface is 15cm then speed of light in the material of lens in 10⁸ms^-1is

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1.5

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answer is B.

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Detailed Solution

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{20} - \frac{1}{- 20} = \frac{1}{f} \Rightarrow f = 10 c m$

$\frac{R}{2 (μ - 1)} = 10 (μ - 1) = \frac{15}{20} μ = \frac{c_{o}}{c_{m}} = \frac{7}{4}$

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