Chapter 20, Problem 70A

To determine

The focal length of the mirror.

Answer to Problem 70A

  $-48\text{cm}$

Explanation of Solution

Given:

The size of the image, $h_i=\frac{3}{4}\left(\text{size of the object}\right)=\frac{3}{4}{h}_o$

The image distance from the mirror, $d_i=-12.0\text{cm}$ (negative sign is used since it is formed behind the mirror)

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 formula of magnification, the object distance will be,

  $\begin{array}{c}\frac{\left(\frac{3}{4}{h}_o\right)}{h_o}=-\frac{\left(-12\right)}{d_o}\\ d_o=16.0\text{cm}\end{array}$

Then using the given values in Mirror equation, the focal length will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{f}=\frac{1}{-12}+\frac{1}{16}\\ f=\frac{16\times \left(12\right)}{12-16}\\ =-48\text{cm}\end{array}$

Negative sign denotes that the mirror is a convex mirror.

Conclusion:

Hence, the focal length of the mirror is $-48\text{cm}$ .