A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left to right as shown in figure(a). If the same lens is instead silvered on the curved surface and illuminated from the other side as shown in figure (B), it acts like an optical system of focal length 10cm. The refractive index of the material of lens is

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1.75

1.51

1.50

1.55

answer is C.

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Detailed Solution

For figure (A)

$\frac{1}{f_{1}} = (\frac{μ - 1}{R}); f = - 28 cm$

$P = 2 P_{1} + P_{2} = 2 P_{1} + 0 or$
$- \frac{1}{28} = - 2 (\frac{μ - 1}{R})$ ……(i)

For figure (B),

$\begin{array}{l} \frac{1}{f_{1}} = (\frac{μ - 1}{R}) f_{2} = - \frac{R}{2} R = - 10 cm \\ P = 2 P_{1} + P_{2} \Rightarrow - \frac{1}{10} = - 2 (\frac{μ - 1}{R}) - \frac{2}{R} \\ \frac{1}{10} = \frac{1}{28} + \frac{2}{R} (using . . (i)) \\ \frac{2}{R} = \frac{1}{10} - \frac{1}{28} = \frac{18}{280}; R = \frac{280}{9} cm \end{array}$

Substituting R in eqn. (i)
$\begin{array}{l} \frac{1}{28} = 2 (\frac{μ - 1}{280}) 9; μ - 1 \\ = \frac{5}{9}; μ = 1 + \frac{5}{9} = \frac{14}{9} = 1.55 \end{array}$