Chapter 17.2, Problem 22PP

a.

To determine

The object’s distance.

Answer to Problem 22PP

  $32\text{cm}$ in front of the mirror.

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 distance of the image from the mirror, $d_i=-24\text{cm}$ (negative sign shows that image is formed behind the mirror).

Formula used:

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

Where,

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

Calculation:

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

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{\left(\frac{3h_o}{4}\right)}{h_o}=-\frac{\left(-24\right)}{d_o}\\ d_o=32\text{}\text{cm}\end{array}$

Negative sign denotes that the image is virtual.

Conclusion:

Hence, the object’s distance is $32\text{cm}$ in front of the mirror.

b.

To determine

The focal length of the mirror.

Answer to Problem 22PP

  $-96\text{cm}$

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 distance of the image from the mirror, $d_i=-24\text{cm}$ (negative sign shows that image is formed behind the mirror)

From part (a), the image distance, $d_o=32\text{}\text{cm}$

Formula used:

Mirror equation,

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

Where,

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

Calculation:

Using the given values in Mirror equation, the focal length of the mirror will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ f=\frac{d_i{d}_o}{d_i+d_o}\\ =\frac{\left(-24\right)\left(32\right)}{-24+32}\\ =-96\text{cm}\end{array}$

Conclusion:

Hence, the focal length of the mirror is $-96\text{cm}$ .