Chapter 17.2, Problem 24SSC

To determine

The magnification of the image.

Answer to Problem 24SSC

The magnification of the image is $-0.82$ .

Explanation of Solution

Given:

The position of the object is $d_{\text{o}}=20.0\text{ cm}$ .

The focal length is $f=9.0\text{ cm}$

Formula used:

The expression for the mirror equation is,

  $\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}$

Here, $d_{\text{o}}$ is the position of the object, $d_{\text{i}}$ is the position of the image, and $f$ is the focal length.

The expression for the magnification $\left(m\right)$ of the image is,

  $m=\frac{-d_{\text{i}}}{d_{\text{o}}}$

Calculation:

The position of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{0}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(9.0\text{ cm}\right)\left(20.0\text{ cm}\right)}{\left(20.0\text{ cm}\right)-\left(\text{9}\text{.0 cm}\right)}\\ d_{\text{i}}=16.4\text{cm}\end{array}$

The magnification of the image is

  $\begin{array}{c}m=\frac{-d_{\text{i}}}{d_{\text{o}}}\\ m=-\frac{16.4cm}{20.0\text{cm}}\\ m=-0.82\end{array}$

Conclusion:

Thus, the magnification of the image is $-0.82$ .