Question

A mirror is part of sphere of diameter 9.3cm a boy stands 0.5m away from the mirror and is amused to see his image. Determine the height of the boy's image if he is 1.3m tall. Where does the boy find his image?

Answer 1

1) Image is at -2.23 cm (virtual)

2) Height of the image: 0.058 m

Explanation:

1)

The location of the image can be found by using the mirror equation:

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

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem, we have:

d = 9.3 cm is the diameter of the mirror, so its radius is

r = 4.65 cm

For a curved mirror, the focal length is half the radius, so

$f=\frac{r}{2}=\frac{4.65}{2}=0.0233 m$

Moreover, the mirror is part of a sphere, so we can assumed it is curved outward; therefore it is a diverging mirror, so its focal length is negative:

$f=-0.0233 m$

The distance of the boy from the mirror is

$p=0.5 m$

So, we can find the distance of the image from the mirror:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-0.0233}-\frac{1}{0.5}=-44.92 m^{-1}$

$q=\frac{1}{-44.92}=-0.0223 m$

So, the image is at -2.23 cm, and it is a virtual image (due to the negative sign)

2)

We can find the height of the image by using the magnification equation:

$M=\frac{y'}{y}=-\frac{q}{p}$

where here we have:

y' = size of the image

y = 1.3 m is the height of the boy

p = 0.5 m is the distance between the boy and the mirror

q = -0.0223 m is the distance between the image and the mirror

And solving for y', we find:

$y'=-\frac{qy}{p}=-\frac{(-0.0223)(1.3)}{0.5}=0.058 m$