Chapter 17.2, Problem 21PP

a.

To determine

The image’s distance.

Answer to Problem 21PP

  $0.48\text{m}$ behind the mirror

Explanation of Solution

Given:

The height of the girl, $h_o=1.8\text{m}$

The distance of the girl from the mirror, $d_o=2.4\text{m}$

The height of the image, $h_i=0.36\text{m}$

Formula used:

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

Where,

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

Calculation:

Using the given values in magnification formula, the image distance will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{0.36}{1.8}=-\frac{d_i}{2.4}\\ d_i=-0.48\text{}\text{m}\end{array}$

Negative sign denotes that the image is virtual.

Conclusion:

Hence, the image distance is $0.48\text{m}$ behind the mirror.

b.

To determine

The focal length of the mirror.

Answer to Problem 21PP

  $-0.6\text{m}$

Explanation of Solution

Given:

The height of the girl, $h_o=1.8\text{m}$

The distance of the girl from the mirror, $d_o=2.4\text{m}$

The height of the image, $h_i=0.36\text{m}$

From part (a), the image distance, $d_i=-0.48\text{}\text{m}$

Formula used:

Mirror equation,

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

Where,

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

Calculation:

Using the given values in Mirror equation, the focal length of the mirror will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ f=\frac{d_i{d}_o}{d_i+d_o}\\ =\frac{\left(-0.48\right)\left(2.4\right)}{-0.48+2.4}\\ =-0.6\text{m}\end{array}$

Conclusion:

Hence, the focal length of the mirror is $-0.6\text{m}$ .