Question

A person has near point at $100\text{cm}$. What radius of curvature of lens is needed to read at $20\text{cm}$ if he/she uses contact lens.
(Refractive index of glass lens $\mu =1.5$)

• A. $100\text{cm}$
• B. $25\text{cm}$
• C. $20\text{cm}$
• D. $45\text{cm}$

Answer 1

The correct option is B $25\text{cm}$

Given,
Near point $=100\text{cm}$
object distance, $u=-20\text{cm}$
Refractive index of lens $\mu =1.5$
object kept at $20\text{cm}$ will be seen clearly, when its image is formed at $100\text{cm}\text{(near point)}$
So,
$u=-20\text{cm}$, $v=-100\text{cm}$
Using lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\Rightarrow \frac{1}{f}=-\frac{1}{100}-\frac{1}{-20}=\frac{1}{25}$
$\therefore f=25\text{cm}$
Now, Using lens maker's formula.
Let $R$ be the radius of curvature of the convex lens.
$\frac{1}{f}=\frac{{\mu }_l-{\mu }_s}{{\mu }_s}(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{25}=\frac{1.5-1}{1}(\frac{1}{R}+\frac{1}{R})$
$\frac{1}{25}=\frac{1}{2}\times \frac{2}{R}$
$R=25\text{cm}$