A plano-convex 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 another side as in fig-$B$, it acts as an optical system of focal length $10 \mathrm{cm}$. The refractive index of the material of the lens is:

• Correct answer
  A

  $1.55$
• B

  $1.50$
• C

  $1.75$
• D

  $1.51$

Solution

Case-1



$\begin{array}{l}\frac{1}{f_1}=\frac{(\mu -1)}{R},f=-28\end{array}$$\Rightarrow P=2p_1+p_2$
$\Rightarrow \frac{1}{28}=2\left(\frac{\mu -1}{R}\right)$

Case-2



$\begin{array}{l}\frac{1}{f_1}=\frac{(\mu -1)}{R}, f_2=\frac{R}{2}\mathrm{ , f }=-10 \mathrm{cm}\end{array}$
$\Rightarrow P=2P_1+P_2\Rightarrow \frac{1}{10}=2\left(\frac{\mu -1}{2}\right)+\frac{2}{R}$$\Rightarrow \frac{1}{10}=\frac{2}{28}+\frac{2}{R}$

$\frac{2}{R}=\frac{1}{10}-\frac{2}{28}=\frac{18}{280}$$\Rightarrow R=\frac{280}{9} \mathrm{cm}$$\Rightarrow \frac{1}{28}=2\left(\frac{\mu -1}{280}\right)9$

$\mu -1=\frac{5}{9}$$\Rightarrow \mu =1+\frac{5}{9}=\frac{14}{9}=1.55$.

Hence, the correct answer is option: 'A'.