# Question #3c1b4

Sep 6, 2015

I found:
Position$= 77 c m$
Real Image bigger and upside down.

#### Explanation:

Consider first the ray diagram:

You can see that the REAL image (formed by the crossing of real rays) will be upside down and quite far from the surface of the mirror.

We now use the mirror equation:
$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ to find the location $q$ of the image:
$\frac{1}{27} + \frac{1}{q} = \frac{1}{20}$
$q = 77 c m$ corresponding to the position of the screen.

The characteristics of the image can be found using the $m$ formula:
$m = - \frac{q}{p} = \frac{h '}{h}$
Where:
$h = 8 c m$ height of object;
$h ' =$ height of image;
$p = 27 c m > 0$
$q = 77 c m > 0$

We get:
$m = - \frac{77}{27} = - 2.85$ this tells us that the image will be upside down ($-$ sign) and bigger (it will be $h ' = 2.85 \cdot 8 \approx 23 c m$ upside down)