Chapter 25, Problem 22P

To determine

The object distance to get maximum magnification for the normal relaxed eye.

Answer to Problem 22P

Solution:

$u=6.17\text{cm}$

$M=4.051\text{X}$

Explanation of Solution

Given:

$\begin{array}{l}f=\text{8}.\text{2}0\text{cm}\\ v=\text{25cm }\left(\text{for normal eye}\right)\end{array}$

Formula used:

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

Calculation:

According to lens equation,

$\begin{array}{l}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\ \frac{1}{u}=\frac{1}{f}-\frac{1}{v}\\ \frac{1}{u}=\frac{1}{0.082}-\frac{1}{0.25}\\ \frac{1}{u}=16.195\\ u=6.17\text{cm}\end{array}$

Maximum Magnification is given by,

$\begin{array}{l}M=-\frac{v}{u}\\ M=-\frac{-0.25}{0.0617}\\ M=4.051\text{X}\end{array}$