Chapter 17, Problem 89A

To determine

The focal length of the mirror.

Answer to Problem 89A

  $-0.67\text{m}$

Explanation of Solution

Given:

The height of the girl, $h_o=1.6\text{m}$

The distance of the girl from the mirror, $d_o=3.2\text{}\text{m}$

The height of the image, $h_i=0.28\text{m}$

Formula used:

Mirror equation,

  $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

Where,

  $\begin{array}{l}f=\text{focal length}\\ d_i=\text{ image distance}\\ d_o=\text{object distance}\end{array}$

And,

Magnification, $\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

Where,

  $\begin{array}{l}{h}_i=\text{image height}\\ h_o=\text{object height}\end{array}$

Calculation:

Using the formula of magnification, the image distance will be,

  $\begin{array}{l}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{0.28}{1.6}=-\frac{d_i}{3.2}\\ d_i=-0.56\text{m}\end{array}$

Using the above and given values in Mirror equation, the focal length will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{f}=\frac{1}{\left(-0.56\right)}+\frac{1}{3.2}\\ f=\frac{3.2\times \left(-0.56\right)}{3.2-0.56}\\ =-0.67\text{m}\end{array}$

Conclusion:

Hence, the focal length of the mirror is $-0.67\text{m}$ .