Question

An object is 10 cm from the mirror, its height is 1 cm and the focal length is 5 cm. What is the image height? (Indicate the object orientation by including the + or - sign with the answer.)
h = ___cm

Answer 1

The image height is -10 cm (the image is upside down)

Explanation:

We can solve the problem by using the mirror equation:

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

where:

f = 5 cm is the focal length of the mirror

p = 10 cm is the distance of the object from the mirror

q is the distance of the image from the mirror

Solving for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\\\rightarrow q= 10 cm$

So, the distance of the image from the mirror is 10 cm.

Now we can find the image height by using the magnification equation:

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

where

y' is the height of the image

y = 1 cm is the height of the object

and using

p = 10 cm

q = 10 cm

We find the size of the image:

$y' = -\frac{qy}{p}=-\frac{(10)(1)}{10}=-10 cm$

where the negative sign indicates that the image is upside down.

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