Chapter 17, Problem 66A

To determine

The distance of image of R from the mirror and its image height.

Answer to Problem 66A

The distance of image of R from the mirror is 10.71 cm.

Height of image is 0.85 cm.

Explanation of Solution

Given:

The diameter of the sphere that act as a convex mirror, $D=40\text{cm}$

The height of the R, $h_o=12\text{cm}$

The distance of the R from the sphere, $d_o=1.5\text{m = 150}\text{}\text{cm}$

Formula used:

The relation between the radius of curvature of mirror $\left(R\right)$ and its focal length $\left(f\right)$ is,

  $f=\frac{R}{2}$

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:

The radius of the sphere will be,

  $R=\frac{D}{2}=\frac{40\text{cm}}{2}=20\text{cm}$

Then, the focal length of the convex mirror will be,

  $f=\frac{R}{2}=\frac{20\text{}\text{}\text{cm}}{2}=10\text{}\text{cm}$

Using the given values in Mirror equation, the image distance will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{10}=\frac{1}{d_i}+\frac{1}{150}\\ d_i=\frac{150\times 10}{150-10}\\ =10.71\text{cm}\end{array}$

Then, the image height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{h_i}{12}=-\frac{10.71}{150}\\ h_i=-0.85\text{cm}\end{array}$

Negative sign denotes inverted image.

Conclusion:

Therefore, the image of R from the sphere is at a distance of $10.71\text{cm}$ and its image height is $0.85\text{cm}$ .