Question

A convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is $(\mu=$ the refractive index of the material of the lens $)$

• A. f
• B. $ \frac{f}{2} $
• C. 2f
• D. $ \left(\mu-1\right) f $

Answer 1

Answer: (C)

The given, $R_{1}=R, R_{2}=-R$
$f=F$

Lens Maker's formula
$\frac{1}{F}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] $
$\frac{1}{f}=(\mu-1)\left[\frac{1}{R}+\frac{1}{R}\right]$
$f=\frac{R}{2(\mu-1)} $
$R=2 f(\mu-1)\,\,\,...(i)$
Now, it is divided vertically into two identical plano convex lens

$\frac{1}{f^{'}}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\left[\because R_{1}=R, R_{2}=\infty\right]$
$\frac{1}{f_{1}}=(\mu-1)\left[\frac{1}{R}-\frac{1}{\infty}\right]$
$f_{1}=\frac{R}{(\mu-1)}$
$f_{1}=\frac{2 f(\mu-1)}{(\mu-1)} $
$f_{1}=2\, f$