# How do you find the position and magnification of a convex mirror?

## Assume the reflected object is $3.00$ cm high and is placed $20.0$ cm from the convex mirror with focal length of $8.00$ cm.

May 13, 2017

See below.
Position: $- 0.05714$ $\text{m}$
Magnification: $0.28571$

#### Explanation:

Since we are given that the height of the object is $0.03$ meters high and is placed $0.2$ meters away from a convex mirror with focal length $- 0.08$ meters, we can write out are givens in SI units as:
${d}_{o b j} = .2$
${h}_{o b j} = .03$ since the image created by convex mirrors are always upright and thus have a positive height value
$f = - 0.08$ since the focal length of convex mirrors are negative

Consider the following two formulas:

Lensmaker's Formula:
$\frac{1}{f} = \frac{1}{d} _ \left(o b j\right) + \frac{1}{d} _ \left(i m g\right)$

Magnification Equation:
$M = {h}_{i m g} / {h}_{o b j} = - {d}_{i m g} / {d}_{o b j}$

To determine the image's position, we can solve for ${d}_{i m g}$ in the Lensmaker's Formula with variables only, then plug in the given values to solve:
$\frac{1}{d} _ \left(i m g\right) = \frac{1}{f} - \frac{1}{d} _ \left(o b j\right)$
Taking the reciprocal of both sides, we get:
d_(img)=1/(1/f-1/d_(obj)

Now, we can substitute the given values of ${d}_{o b j}$ and $f$ to solve:
${d}_{i m g} = \frac{1}{\frac{1}{-} 0.08 - \frac{1}{0.2}}$
$= - 0.05714$ $\text{m}$

Using this value, we can find the magnification of the convex mirror:
$M = - {d}_{i m g} / {d}_{o b j}$
$= - \frac{- 0.05714}{0.2}$
$= 0.28571$ which has no units

May 13, 2017

See below.

#### Explanation:

You will need to calculate the distance between the mirror and the image first, which can be done using the mirror equation:

$\frac{1}{f} = \frac{1}{d} _ \left(o\right) + \frac{1}{d} _ i$

where $f$ is the focal length, ${d}_{o}$ is the distance between the mirror and the object, and ${d}_{i}$ is the distance between the mirror and the image.

We can solve for ${d}_{i}$:

$\implies \frac{1}{d} _ i = \frac{1}{f} - \frac{1}{d} _ \left(o\right)$

$\implies {d}_{i} = {\left(\frac{1}{f} - \frac{1}{d} _ \left(o\right)\right)}^{-} 1$

Note that because this is a convex mirror, the focal length must be negative.

Given that ${d}_{o} = 20.0 c m$ and $f = - 8.00 c m$ :

${d}_{i} = {\left(- \frac{1}{8} - \frac{1}{20}\right)}^{-} 1$

$= {\left(- \frac{7}{40}\right)}^{-} 1$

$= - \frac{40}{7} c m$

The magnification of a curved mirror can be expressed by the following equation:

$m = - {d}_{i} / {d}_{o}$

Thus we have:

$m = \frac{- \left(- \frac{40}{7}\right)}{20}$

$m = \frac{40}{140} = \frac{2}{7}$

$\therefore$ The position of the image is $\frac{40}{7}$ cm behind the mirror and the magnification of the mirror is $\frac{2}{7}$.

This answer makes sense, as a convex mirror will always produce an image which is reduced, upright, and virtual.