Chapter 17, Problem 80A

To determine

The image position and the image height.

Answer to Problem 80A

The image position is 23 cm in the front of the mirror and its height is 1.84 cm.

Explanation of Solution

Given:

The radius of curvature of the concave mirror, $R=26.0\text{}\text{cm}$

The height of the object, $h_o=2.4\text{cm}$

The object distance from the mirror, $d_o=30.0\text{cm}$

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:

The focal length of the concave mirror will be,

  $f=\frac{26.0}{2}=13.0\text{cm}$

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

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{13}=\frac{1}{d_i}+\frac{1}{30}\\ d_i=\frac{30\times 13}{30-13}\\ \approx 23\text{cm}\end{array}$

Then, the image height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{2.4}=-\frac{23}{30.0}\\ h_i=-1.84\text{cm}\end{array}$

Negative sign denotes inverted image.

Conclusion:

Therefore, the image is formed at a distance of $23\text{}\text{cm}$ in the front of the mirror and its height is $1.84\text{cm}$ .