Chapter 18.2, Problem 15PP

To determine

The image position and image height.

Answer to Problem 15PP

  $15.6\text{cm}$ to the right of the lens and $4.13\text{cm}$ , respectively.

Explanation of Solution

Given:

The focal length of the convex lens, $f=5.5\text{cm}$

The height of the object, $h_o=2.25\text{cm}$

The object distance from the lens, $d_o=8.5\text{cm}$

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,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{5.5}=\frac{1}{d_i}+\frac{1}{8.5}\\ d_i=\frac{8.5\times 5.5}{8.5-5.5}\\ =15.6\text{cm}\end{array}$

Then, the image height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{2.25}=-\frac{15.6}{8.5}\\ h_i=-4.13\text{cm}\end{array}$

Negative sign denotes inverted image.

Conclusion:

Hence, the image position is $15.6\text{cm}$ to the right of the lens and image height is $4.13\text{cm}$ .