Chapter 18.2, Problem 24SSC

To determine

The image position, image height and its orientation.

Answer to Problem 24SSC

The image distance is $3.0\text{m}$ to the right of the lens, is real and inverted, and is $4.0\text{m}$ in height.

Explanation of Solution

Given:

The focal length of the convex lens, $f=1.0\text{m}$

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

The object distance from the lens, $d_o=1.5\text{m}$

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,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{1.0}=\frac{1}{d_i}+\frac{1}{1.5}\\ d_i=\frac{1.5\times 1.0}{1.5-1.0}\\ =3\text{m}\end{array}$

Then, the image height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{2.0}=-\frac{3.0}{1.5}\\ h_i=-4\text{m}\end{array}$

Negative sign denotes that the image is inverted and real.

Conclusion:

Hence, the image is formed at a distance of $3\text{m}$ to the right of the lens, is real and inverted, and is $4\text{m}$ tall.