Chapter 17, Problem 6STP

To determine

To Choose: The correct option.

Answer to Problem 6STP

  $23\text{cm}$

Explanation of Solution

Given:

The magnification of the image, $m=+3.2$

The object distance, $d_o=30\text{cm}$

Formula used:

Mirror equation,

  $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

Where,

  $\begin{array}{l}f=\text{focal length}\\ d_i=\text{ image distance}\\ d_o=\text{object distance}\end{array}$

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 formula of magnification, the image distance will be,

  $\begin{array}{c}m=-\frac{d_i}{d_o}\\ +3.2=-\frac{d_i}{d_o}\\ d_i=-3.2d_o\end{array}$

Then using Mirror Equation, the focal length will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{f}=\frac{1}{3.2d_o}+\frac{1}{d_o}\\ f=\frac{d_o\left(3.2\right)}{3.2+1}\\ =\frac{\left(30\right)\left(3.2\right)}{4.2}\\ \approx 23\text{cm}\end{array}$

Conclusion:

Hence, the focal length of the mirror is $23\text{cm}$ .