Chapter 25, Problem 11P

To determine

The power of the reading glasses.

Answer to Problem 11P

Solution:

Power = 1.068D

Explanation of Solution

Given:

$\begin{array}{l}\text{Lens}-\text{eye distance is 1}.\text{8 cm}.\\ u=\text{object distance}=-\text{123}.\text{2cm}=-\text{1}.\text{232m}\\ v=\text{image distance}=\text{53}.\text{2cm}=0.\text{532m}\end{array}$

Formula used:

$\begin{array}{l}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\ P=\frac{1}{f}\end{array}$

Calculation:

According to lens equation,

$\begin{array}{l}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\ \frac{1}{f}=\frac{1}{0.532}-\frac{1}{1.232}\\ \frac{1}{f}=1.8796-0.8116\\ \frac{1}{f}=1.068\\ \text{Power of the reading glass is given by,}\\ P=\frac{1}{f}\\ P=1.068D\end{array}$