A biconvex lens is made of glass with refractive index 1.5 and radii of curvature 20 cm and 30 cm. if the surface whose radius is 20 cm silvered, the effective focal length of the optical device formed is

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$- \frac{69}{11} cm$

$- \frac{79}{11} cm$

$- \frac{60}{11} cm$

$\frac{60}{11} cm$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{f_{eq}} = \frac{1}{f_{L}} + \frac{1}{f_{m}} + \frac{1}{f_{L}} and \\ \frac{1}{f_{L}} = (μ - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}}) = (1 .5 - 1) (\frac{1}{20} + \frac{1}{30}) \\ = \frac{1}{2} (\frac{30 + 20}{20 \times 30}) = \frac{1}{2} \times \frac{50}{20 \times 30} = \frac{5}{120}; \\ \frac{1}{f_{eq}} = \frac{5}{120} + \frac{2}{20} + \frac{5}{120} = \frac{5 + 12 + 5}{120} = \frac{22}{120} \\ f_{eq} = \frac{60}{11} cm \\ as it works as a concave mirror , f_{eq} = - \frac{60}{11} cm \end{array}$