Chapter 17.2, Problem 20PP

To determine

To compute: - The position of image of the ball and its image diameter.

Answer to Problem 20PP

  $u\text{ = 12}\text{.69 cm}$

  $d=h_i=\text{4}\text{.38 cm}$

Explanation of Solution

Given info:

Object distance, $u=-22\text{ cm =}-0.22\text{ m}$

Radius of curvature, $R=+60\text{ cm = +0}\text{.6 m}$

Ball’s diameter = Height of object, $h_o=7.6\text{ cm = 0}\text{.076 m}$

Formula used:

The relation between object distance $\left(u\right)$ , image distance $\left(v\right)$ , and radius of curvature $\left(R\right)$ can be given by:

  $\frac{1}{u}+\frac{1}{v}=\frac{2}{R}$

Calculation:

Plugging in the given values in the formula to calculate the position of image:

  $\begin{array}{c}\frac{1}{-0.22}+\frac{1}{v}=\frac{2}{0.6}\\ \therefore \\ \frac{1}{v}=\frac{2}{0.6}+\frac{1}{0.22}\\ \frac{1}{v}=\frac{1.04}{0.132}\\ \therefore \\ v=0.1269\text{ m}\\ \text{= 12}\text{.69 cm}\end{array}$

Positive value of $v$ shows that the image is formed beyond the mirror.

The nature of image is virtual erect.

Magnification can be defined as:

  $m=\frac{h_i}{h_o}=-\frac{v}{u}$

Therefore, Height of image:

  $h_i=-\frac{v}{u}\times h_o$

Plugging in the given values:

  $\begin{array}{c}{h}_i=-\frac{0.1269}{-0.22}\times 0.076\\ =0.0438\text{ m}\\ \text{= 4}\text{.38 cm}\end{array}$

Diameter of the image of the ball, $d=h_i=\text{4}\text{.38 cm}$

Conclusion:

Position of image, $v=12.69\text{ cm}$

Diameter of the image of the ball, $d=h_i=\text{4}\text{.38 cm}$