Question

A plano-convex lens, when silvered in the plane side, behaves like a concave mirror of focal length $30cm$. When it is silvered on the convex side, it behaves like a concave mirror of focal length $10cm$. The refractive index of the material of the lens is :

• A. $1.33$
• B. $1.5$
• C. $1.55$
• D. $1.6$

Answer 1

The correct option is B $1.5$

$\begin{array}{c}\text{let the radius of the convex side}=R\\ \text{inverse focal length of lens}=(\mu -1)(\frac{1}{\infty }-\frac{1}{-R})\\ \begin{array}{cc}\frac{1}{f_R}& =\frac{\mu -1}{R}.\\ \text{for finding focal length of a silvered lens}& \\ \frac{1}{f}& =\frac{1}{f_m}-\frac{2}{f_L}\\ f^{}& =\text{Resultant focal length}\\ f_m& =\text{focal length of mirror}\\ f_L& =\text{focal length of lens}\end{array}\end{array}$

$\begin{array}{c}\text{case}-1\\ \frac{1}{-30}=\frac{1}{\infty }-2\left[\frac{\mu -1}{R}\right]\\ \frac{1}{30}=\frac{2[\mu -1]}{R}-\left(1\right)\\ \text{case}-2\\ \frac{1}{-10}=\frac{1}{-\frac{R}{2}}-2\left[\frac{\mu -1}{R}\right]\\ \frac{1}{10}=\frac{2[\mu -1+1]}{R}\\ \frac{R}{M}=20\Rightarrow R=20\mu -\left(2\right)\\ \text{Substitute (2) in (1)}\end{array}$

$\begin{array}{cc}\frac{1}{30}& =\frac{2(\mu -1)}{20\mu }\\ \frac{\mu }{3}& =\mu -1\\ 1& =\frac{2\mu }{3}\\ & \mu =\frac{3}{2}\Rightarrow \mu =1.5\end{array}$

Hence option (B) is correct