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)

2f

Answer 2

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