Chapter 17, Problem 68A

To determine

The distance of watch’s image from the mirror and its image’s diameter.

Answer to Problem 68A

The watch’s image is 24 cm behind the mirror.

Diameter of image is 9.0 cm.

Explanation of Solution

Given:

The focal length of the concave mirror, $f=12.0\text{cm}$

The diameter of the watch, $h_o=3.0\text{cm}$

The watch’s distance from the mirror, $d_o=8.0\text{cm}$

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 given values in Mirror equation, the watch’s image distance will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{12}=\frac{1}{d_i}+\frac{1}{8.0}\\ d_i=\frac{8\times 12}{8-12}\\ =-24\text{cm}\end{array}$

Negative sign denotes that image is formed behind the mirror.

Then, the diameter of watch’s image will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{3.0}=-\frac{\left(-24\right)}{8.0}\\ h_i=9.0\text{cm}\end{array}$

Conclusion:

Therefore, the image of watch is formed at a distance of $24\text{}\text{cm}$ behind the mirror and its image’s diameter is $9.0\text{cm}$ .