Question 201 of 295

A plano-convex lens becomes an optical system of $28cm$ 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 $10cm$. The refractive index of the material of the lens is:

A$1.55$B$1.50$C$1.75$D$1.51$

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Answer: (A) $1.55$

Case-1



$\begin{matrix}\dfrac{1}{{f}_{1}}=\dfrac{(\mu -1)}{R},f=-28\end{matrix}$$\Rightarrow P=2{p}_{1}+{p}_{2}$
$\Rightarrow \dfrac{1}{28}=2\left(\dfrac{\mu -1}{R}\right)$

Case-2



$\begin{matrix}\dfrac{1}{{f}_{1}}=\dfrac{(\mu -1)}{R},{f}_{2}=\dfrac{R}{2}, f=-10cm\end{matrix}$
$\Rightarrow P=2{P}_{1}+{P}_{2}\Rightarrow \dfrac{1}{10}=2\left(\dfrac{\mu -1}{2}\right)+\dfrac{2}{R}$$\Rightarrow \dfrac{1}{10}=\dfrac{2}{28}+\dfrac{2}{R}$

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

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