Question

A millipede sits $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$in front of the nearest part of the surface of a shiny sphere of diameter $\mathbf{0}\mathbf{.}\mathbf{70}\mathbf{}\mathit{m}$(a) How far from the surface does the millipede’s image appear? (b) If the millipede’s height is $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{m}\mathbf{,}$ what is the image height? (c) Is the image inverted?

Short answer

1. Distance of image from the surface, $i=0.15m$
2. Height of image {H_i} = 0.30m
3. Nature of image $m\le 0$is upright

Step-by-step solution

Determine the given quantities

Diameter of sphere is$D=0.70m$

Height of the millipede is $H_p=2.0m$

Distance ofmillipede is$p=1.0m$

Determine the concepts of optics

As the outer surface of the sphere is shiny, note that it is a convex mirror and using the mirror equation considering the negative focal length, we will get the required answer.

Focal length of the spherical mirror is $f=\frac{R}{2}$as:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$

Magnification$m=-\frac{i}{p}=\frac{H_i}{H_p}$

(a) Determine the distance of image from the surface 

Diameter of sphere is $D=0.70m$

Then radius of sphere will be $R=0.35m$

Focal length of spherical mirror is$=\frac{R}{2}=0.175$

The mirror is convex. Therefore, $f=-0.175$

$\begin{array}{l}\frac{1}{p}+\frac{1}{i}=\frac{1}{f}\\ \frac{1}{1.0m}+\frac{1}{0.175m}=\frac{-1}{i}\\ -\frac{1}{i}=6.7142\\ i\approx -0.15m\end{array}$

Distance of image from the surface,$i=0.15m$

(b) Calculate the height of image.

Consider the formula for the magnification as:

$\begin{array}{l}m=\frac{i}{p}\\ =-\left(\frac{-115}{1}\right)\\ =0.15\end{array}$

Consider the condition as:

$m>0$

$\begin{array}{c}m=\frac{H_i}{H_p}\\ =\frac{H_i}{2.0}\end{array}$

Solve further as:

$\begin{array}{c}{H}_i=0.15\times 2\\ =0.30m\end{array}$

Height of image$H_i=0.30m$

(c) Determine if the image is inverted or not 

Consider the value of the magnification$m=0.15>0$

The image will be upright or not inverted (NI).