Chapter 17.2, Problem 29SSC

To determine

The focal length of the mirror.

Answer to Problem 29SSC

The focal length of the mirror is $-36\text{ cm}$ .

Explanation of Solution

Given:

The magnification is $m=\frac{2}{3}$ .

The position of the image is $d_{\text{i}}=-12.0\text{ cm}$ .

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 equation is,

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

Here, $h_{\text{i}}$ is the height of the image and $h_{\text{o}}$ is the height of the object.

Calculation:

The position of the object is

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ d_{\text{o}}=-\frac{d_{\text{i}}}{m}\\ d_{\text{o}}=-\frac{\left(-12.0\text{ cm}\right)}{\left(\frac{2}{3}\right)}\\ d_{\text{o}}=18.0\text{ cm}\end{array}$

The focal length of the mirror is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ f=\frac{d_{\text{o}}{d}_{\text{i}}}{d_{\text{i}}+d_{\text{o}}}\\ f=\frac{\left(18.0\text{ cm}\right)\left(-12.0\text{ cm}\right)}{\left(-12.0\text{ cm}\right)+\left(18.0\text{ cm}\right)}\\ f=-36.0\text{ cm}\end{array}$

Conclusion:

Thus, the focal length of the mirror is $-36.0\text{ cm}$ .