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. $(\mu -1)f$

Answer 1

Answer: (C)

Using lens-maker formula : $\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

For convex lens : $R_1=R$ $R_2=-R$

$\therefore$ $\frac{1}{f}=(\mu -1)(\frac{1}{R}-\frac{1}{-R})$ $⟹$ $\frac{1}{f}=(\mu -1)\frac{2}{R}$ ........(1)

For plano-convex lens : $R_1=R$ $R_2=\infty$

$\therefore$ $\frac{1}{f^{\prime }}=(\mu -1)(\frac{1}{R}-\frac{1}{\infty })$ $⟹$ $\frac{1}{f^{\prime }}=(\mu -1)\frac{1}{R}=\frac{1}{2f}$

$⟹$ $f^{\prime }=2f$

Thus the focal length of Plano-convex lens is $2f$.