Chapter 17, Problem 3STP

To determine

To Choose: The correct option.

Answer to Problem 3STP

  $-34\text{cm}$

Explanation of Solution

Given:

The height of the image, $h_i=\frac{3}{4}\left(\text{height of the object}\right)=\frac{3h_o}{4}$

The image distance, $d_i=-8.4\text{}\text{cm}$ (Negative sign denotes that the image is formed behind the mirror)

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, $\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

Where,

  $\begin{array}{l}{h}_i=\text{image height}\\ h_o=\text{object height}\end{array}$

Calculation:

Using the given values in magnification formula, the object distance will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{\left(\frac{3h_o}{4}\right)}{h_o}=-\frac{\left(-8.4\right)}{d_o}\\ d_o=11.2\text{}\text{cm}\end{array}$

Then using the Mirror equation, the focal length of the mirror will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{f}=\frac{1}{-8.4}+\frac{1}{11.2}\\ f=\frac{11.2\times \left(-8.4\right)}{-11.2+8.4}\\ \approx -34\text{cm}\end{array}$

Conclusion:

Hence, the focal length of the mirror is $-34\text{cm}$ .