Chapter 19, Problem 73A

To determine

The image position and the image height.

Answer to Problem 73A

The image position is $234\text{cm}$ behind the mirror and its height is $4\text{cm}$ .

Explanation of Solution

Given:

The radius of curvature of the concave mirror, $R=48.0\text{}\text{cm}$

The height of the object, $h_o=2.0\text{cm}$

The object distance from the mirror, $d_o=12.0\text{cm}$

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

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,

  $f=\frac{48.0}{2}=24.0\text{cm}$

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

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

Negative sign denotes that the image is formed behind the mirror.

Then, the image height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{2.0}=-\frac{\left(-24.0\right)}{12.0}\\ h_i=4\text{cm}\end{array}$

Conclusion:

Hence, the image is formed at a distance of $234\text{cm}$ behind the mirror and its height is $4\text{cm}$ .