Chapter 9, Problem 44Q

To determine

To estimate:

The value of the magnification shown in figures 9.45, 9.48, 9.49.

Explanation of Solution

Given:

Figure 9.45

Figure 9.48

Figure 9.49

Formula used:

Magnification, $m=\frac{v}{u}$, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\begin{array}{l}where\\ mismagnification\\ visimageheight\\ uisobjectheight\end{array}$.

Calculation:

For figure 9.45:

$\begin{array}{l}v=p\\ u=-s\\ m=-\frac{p}{s}=1-\frac{p}{f}\\ s>p,somagnificationislessthan1\end{array}$ $\begin{array}{l}valueofmagnification\\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{p}-\frac{1}{-s}=\frac{1}{f}\\ 1+\frac{p}{s}=\frac{p}{f}\Rightarrow \frac{p}{s}=\frac{p}{f}-1\end{array}$

For figure 9.48:

$\begin{array}{l}v=-p\\ u=-s\\ m=\frac{-p}{-s}=\frac{p}{s}=\frac{p}{f}+1\\ s<p,somagnificationisgreaterthan1\end{array}$ $\begin{array}{l}valueofmagnification\\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{-p}-\frac{1}{-s}=\frac{1}{f}\\ -1+\frac{p}{s}=\frac{p}{f}\Rightarrow \frac{p}{s}=\frac{p}{f}+1\end{array}$

For figure 9.49:

$\begin{array}{l}v=-p\\ u=-s\\ m=\frac{-p}{-s}=\frac{p}{s}=1-\frac{p}{f}\\ s>p,somagnificationislessthan1\end{array}$ $\begin{array}{l}valueofmagnification\\ \frac{1}{v}+\frac{1}{u}=-\frac{1}{f}\\ \frac{1}{-p}+\frac{1}{-s}=-\frac{1}{f}\\ -1+\frac{p}{s}=-\frac{p}{f}\Rightarrow \frac{p}{s}=1-\frac{p}{f}\end{array}$.