Chapter 17, Problem 63A

To determine

The magnification of the image.

Answer to Problem 63A

The magnification of the image is 5.

Explanation of Solution

Given:

The radius of curvature of the concave mirror, $R=40\text{mm}$

The distance of tooth from the mirror, $d_o=16.0\text{mm}$

Formula used:

The relation between the radius of curvature of mirror $\left(R\right)$ and its focal length $\left(f\right)$ is,

  $f=\frac{R}{2}$

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

Where,

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

Calculation:

Then, the focal length of the concave mirror will be,

  $f=\frac{R}{2}=\frac{40\text{}\text{}\text{mm}}{2}=20\text{}\text{mm}$

Using the given values in Mirror equation, the image distance will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{20}=\frac{1}{d_i}+\frac{1}{16}\\ d_i=\frac{16\times 20}{16-20}\\ =-80\text{mm}\end{array}$

Then, the magnification of the image will be,

  $\begin{array}{c}m=-\frac{d_i}{d_o}\\ =-\frac{\left(-80\right)}{16}\\ =5\end{array}$

Conclusion:

Therefore, the magnification of the image is $5$ .