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
$\frac{1}{F}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$\frac{1}{f}=(\mu -1)\left[\frac{1}{R}+\frac{1}{R}\right]$
$f=\frac{R}{2(\mu -1)}$
$R=2f(\mu -1)...(i)$
Now, it is divided vertically into two identical plano convex lens

$\frac{1}{f^{{}^{\prime }}}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]\left[\because R_1=R,R_2=\infty \right]$
$\frac{1}{f_1}=(\mu -1)\left[\frac{1}{R}-\frac{1}{\infty }\right]$
$f_1=\frac{R}{(\mu -1)}$
$f_1=\frac{2f(\mu -1)}{(\mu -1)}$
$f_1=2f$