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A thin equiconvex lens has focal length 10 cm and refractive index 1.5. One of its faces is now silvered, and for an object placed at a distance u infront of it, the image coincides with the object. The value of u is

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10 cm

5 cm

20 cm

15 cm

answer is B.

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Detailed Solution

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}$

$\frac{1}{F} = \frac{2 (μ - 1)}{R} + \frac{2 (μ - 1)}{R} + \frac{2}{R}; then \frac{1}{F} = \frac{1}{v} - \frac{1}{u}$

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