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A plano–convex lens $(μ = 1.5)$ of focal length 20cm has its plane side silvered:

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It acts as a convex mirror

The radius of curvature of its curved surface is half that of a surface of equi-convex lens of focal length 20cm made of same material

An object placed at a distance of 15cm on the axis of the convex side gives rise to an image at a distance of 30cm from it

An object placed at a distance of 20cm on the axis of the convex side gives rise to an image at a distance of 40cm from it

answer is A, B.

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Detailed Solution

For the given plano-convex lens (without silvering),

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

$\frac{1}{20} = (1.5 - 1) (\frac{1}{R} - \frac{1}{\infty}) R = 10 c m$

For equiconvex lens,

$\frac{1}{20} = (1.5 - 1) (\frac{2}{R}) R = 20 c m$

Hence, option A is correct
For the silvered lens,
$\frac{1}{F} = \frac{1}{f_{m}} - \frac{2}{f_{l}} = \frac{1}{\infty} - 2 (1.5 - 1) (\frac{1}{10} - \frac{1}{\infty})$

$F = - 10 cm$ (concave mirror)

$\frac{1}{v} + \frac{1}{u} = \frac{1}{F} \Rightarrow \frac{1}{v} + \frac{1}{- 15} = \frac{1}{- 10}$

$\frac{1}{v} = \frac{1}{15} - \frac{1}{10} v = - 30 c m$

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