A biconvex thin lens is prepared from glass of refractive index 3/2. The two bounding surfaces have equal radii of 25 cm each. One of the surfaces is silvered from outside to make it reflecting. Where should an object be placed before this lens so that the image coincides with the object.

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Detailed Solution

Here, $R_{1} = + 25 cm, R_{2} = - 25 cm, and μ = 3 / 2$
Image coincides with object, hence, u = v = –x (say)
$\frac{1}{F} = \frac{2}{F_{L}} + \frac{1}{F_{C}} = 2 (\frac{3}{2} - 1) \frac{2}{25} + \frac{2}{25} \Rightarrow \frac{1}{F} = \frac{4}{25}$
By using $\frac{1}{V} - \frac{1}{u} = \frac{1}{F} \Rightarrow - \frac{1}{x} - \frac{1}{x} = \frac{4}{25}$ ; x = 12.5 cm
Hence, the object should be placed at a distance 12.5 cm in front of the silvered lens.

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