Question 49 of 295

A planoconvex lens becomes an optical system of $28 \mathrm{~cm}$ focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A. If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig. B, it acts like an optical system of focal length $10 \mathrm{~cm}$. The refractive index of the material of lens is

A1.50B1.55C1.75D1.51

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Answer: (B) 1.55

Case-1


$\frac{1}{f_1}=\left(\frac{\mu-1}{R}\right) \mathrm{f}=-28$
$\mathrm{P}=2 \mathrm{P}_1+\mathrm{P}_2 \Rightarrow \frac{1}{28}=2\left(\frac{\mu-1}{R}\right)$
$\left(\because\right.$ Power, $\left.P=\frac{1}{f} \& f_{\text {plane mirror }}=\infty\right)$
Case-2

$\frac{1}{f_1}=\left(\frac{\mu-1}{R}\right) \quad f_2=-\frac{R}{2} \quad f=-10 \mathrm{~cm}$
$\mathrm{P}=2 \mathrm{P}_1+\mathrm{P}_2 \Rightarrow \frac{1}{10}=2\left(\frac{\mu-1}{2}\right)+\frac{2}{R}$
or, $\quad \frac{1}{10}=\frac{1}{28}+\frac{2}{R} \Rightarrow \frac{2}{R}=\frac{1}{10}-\frac{2}{28}=\frac{18}{280}$
or, $\quad \mathrm{R}=\frac{280}{9} \mathrm{~cm}$
or, $\quad \frac{1}{28}=2\left(\frac{\mu-1}{280}\right) 9 \quad \Rightarrow \quad \mu-1=\frac{5}{9}$
$\therefore \quad \mu=1+\frac{5}{9}=\frac{14}{9}=1.55$