Chapter 19, Problem 74A

To determine

The image position, and image height.

Answer to Problem 74A

The image is formed $11.6\text{cm}$ to the left of the lens, and is $3.09\text{cm}$ tall.

Explanation of Solution

Given:

The focal length of the convex lens, $f=21.0\text{cm}$

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

The candle’s distance from the lens, $d_o=7.50\text{cm}$

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,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{21.0}=\frac{1}{d_i}+\frac{1}{7.5}\\ d_i=\frac{7.5\times 21.0}{7.5-21.0}\\ =-11.6\text{cm}\end{array}$

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

Then, the image height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{2.0}=-\frac{\left(-11.6\right)}{7.50}\\ h_i=3.09\text{cm}\end{array}$

Positive sign denotes that the image is virtual and upright.

Conclusion:

Hence, the image is formed at a distance of $11.6\text{cm}$ to the left of the lens, and is $3.09\text{cm}$ tall.