Chapter 20, Problem 71A

To determine

The image position, image height and its orientation.

Answer to Problem 71A

The image is formed at a distance of $10.9\text{cm}$ to the left of the lens, is virtual and upright, and is $1.09\text{cm}$ tall.

Explanation of Solution

Given:

The focal length of the diverging lens, $f=-24.0\text{cm}$

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

The object distance from the lens, $d_o=20.0\text{cm}$

Formula used:

Thin lens equation,

  $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

Where,

  $\begin{array}{l}f=\text{focal length of the lens}\\ d_i=\text{ image distance}\\ d_o=\text{object distance}\end{array}$

And,

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

Where,

  $\begin{array}{l}{h}_i=\text{image height}\\ h_o=\text{object height}\end{array}$

Calculation:

Using the given values in thin lens equation, the image distance will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{-24}=\frac{1}{d_i}+\frac{1}{20}\\ d_i=-\left(\frac{20\times 24}{20+24}\right)\\ =-10.90\text{cm}\end{array}$

Negative sign denotes that the image is formed to the left of the lens.

Then, the image height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{2.0}=-\frac{\left(-10.90\right)}{20}\\ h_i=1.09\text{cm}\end{array}$

Positive sign denotes that the image is virtual and upright.

Conclusion:

Hence, the image is formed at a distance of $10.9\text{cm}$ to the left of the lens, is virtual and upright, and is $1.09\text{cm}$ tall.