Question

A person has a near point at $100\text{cm}$. The radius of curvature of the lens required so that a person can see objects lying at $20\text{cm}$ clearly is $5n\text{cm}$. The value of $n$ is
(Refractive index of glass lens$=1.5$)

Answer 1

Object kept at $20\text{cm}$ will be seen clearly, when its image is formed at $100\text{cm}$ (near point).

So, $u=-20\text{cm},v=-100\text{cm}$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\Rightarrow \frac{1}{f}=\frac{1}{-100}-\frac{1}{-20}=\frac{1}{25}$

$\therefore f=25\text{cm}$

Let $R$ be the radius of curvature of the convex lens. Using lens maker formula,

$\frac{1}{f}=\frac{{\mu }_l-{\mu }_s}{{\mu }_s}(\frac{1}{R_1}-\frac{1}{R_2})$

$\Rightarrow \frac{1}{25}=\frac{1.5-1}{1}(\frac{1}{R}+\frac{1}{R})$

$\Rightarrow \frac{1}{25}=\frac{1}{2}\times \frac{2}{R}$

$\Rightarrow R=25\text{cm}$

Given, $R=5n\text{cm}$

$\therefore n=5$

Note: When the nature of the convex lens is not mentioned in the question, consider the lens as an equi-convex lens.