Chapter 18, Problem 64A

To determine

The focal length of the lens.

Answer to Problem 64A

  $10.3\text{cm}$

Explanation of Solution

Given:

The height of the image, $h_i=0.75\left(\text{height of the object}\right)=-0.75h_o$ (negative sign denotes that the image is real and inverted)

The distance of drawing from the lens, $d_o=24\text{cm}$

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 magnification formula, the image distance will be,

  $\begin{array}{c}\frac{-0.75h_o}{h_o}=-\frac{d_i}{24\text{cm}}\\ d_i=18\text{cm}\end{array}$

Then, using the given values in Thin lens equation, the focal length will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{f}=\frac{1}{18}+\frac{1}{24}\\ f=\frac{18\times 24}{24+18}\\ =10.3\text{cm}\end{array}$

Conclusion:

Hence, the focal length of the lens is $10.3\text{cm}$ .