Chapter 17, Problem 77A

To determine

To Complete: The given problem.

Explanation of Solution

Given info:

The object distance is, $u=32\text{cm}$

Formula used:

The mirror’s formula is given as,

  $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Where, $f$ represents the focal length, $u$ represents the object distance and $v$ represents the image distance.

Calculation:

The distance is 32cm. Now using the value of image distance and object distance in the above formula and the focal length will be given as,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}=\frac{1}{32}+\frac{1}{v}\\ \frac{1}{f}-\frac{1}{32}=\frac{1}{v}\\ \frac{1}{v}=\frac{32-f}{32f}\end{array}$

To have a virtual image, $\frac{1}{v}$ should be negative.

  $\begin{array}{l}\frac{32-f}{32f}<0\\ 32-f<0\\ f>32\text{cm}\end{array}$

Thus, the focal length should be more than 32 cm such that virtual image will be formed.

The complete question can be:

An object is placed 32 cm from a mirror. Find the focal length such that a virtual image will be formed.