Question
A thin biconvex lens is prepared from the glass ( $\mu = 1.5$ ) both curved surfaces of which have equal radii of 20 cm each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of $\_\_\_\_$ cm.
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Solution
Sol. [Key Concept: A silvered biconvex lens acts as a mirror of power $\frac{1}{f} = \frac{2}{f_{L}} + \frac{1}{f_{m}}$; the object coincides with its image when placed at the equivalent centre of curvature, $u = 10$ cm.]
Combination of lens + mirror
$${f_{m} = \frac{R}{2} = - 10\text{ }cm;f_{L} = \frac{R}{2(\mu - 1)} = 20\text{ }cm }{\frac{1}{f} = \frac{1}{f_{m}} - \frac{2}{f_{L}} }{\frac{1}{f} = \frac{- 1}{10} - \frac{2}{20} = - \frac{1}{5} }{f = - 5\text{ }cm }$$To form image at the position of object, it should be placed at center of curvature of mirror + lens combination.
$$\therefore u = 2f = 2 \times 5 = 10\text{ }cm $$
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