Question #c8fa6

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Question #c8fa6

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Answer 1 · 2017-12-26T17:18:09

See below.

Explanation:

The focal length of a mirror $f$ $f$ is defined as the distance from the mirror to the focal point F, which is the point midway between the vertex (geometric center of mirror) and center of curvature.

If the object is placed 9 cm in front of the mirror and 4 cm from the focal point (I assume), that indicates that the focal point is located in front of the mirror, making this a concave mirror. The focal length is then 5 cm.

The mirror equation (also used for thin lenses) is given by:

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d_{i}}$ $1/f=1/d_"o"+1/d_i$

where $d_{o}$ $d_o$ is the distance from the mirror to the object and $d_{i}$ $d_i$ is the distance from the mirror to the image.

We are given:

$d_{o} = 9 cm$ $d_o=9"cm"$
$f = 5 cm$ $f=5"cm"$
$h_{o} = 4 cm$ $h_o=4"cm"$

We can begin by finding $d_{i}$ $d_i$ with a bit of algebra.

$\Rightarrow \frac{1}{d_{i}} = \frac{1}{f} - \frac{1}{d_{o}}$ $=>1/d_i=1/f-1/d_o$

$\Rightarrow d_{i} = {(\frac{1}{f} - \frac{1}{d_{o}})}^{- 1}$ $=>d_i=(1/f-1/d_o)^-1$

Using our known values:

$d_{i} = {(\frac{1}{5} - \frac{1}{9})}^{- 1}$ $d_i=(1/5-1/9)^-1$

$= {(\frac{9}{45} - \frac{5}{45})}^{- 1}$ $=(9/45-5/45)^-1$

$= {(\frac{4}{45})}^{- 1}$ $=(4/45)^-1$

$= \frac{45}{4}$ $=45/4$

So $d_{i} = \frac{45}{4} cm$ $d_i=45/4"cm"$ or $\approx 11 cm$ $~~11"cm"$

Since the mirror is located at 9 cm from the object, this indicates the the image is formed behind the object and in front of the mirror. This means that the image is real.

Now that we know $d_{i}$ $d_i$ , we can use this to find the magnification m of the mirror. We can then use the magnification to find the image height $h_{i}$ $h_i$ .

$m = \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}}$ $m=h_i/h_o=-d_i/d_o$

$m = - \frac{d_{i}}{d_{o}}$ $m=-d_i/d_o$

$= - \frac{\frac{45}{4}}{9}$ $=-(45/4)/9$

$= - \frac{45}{36}$ $=-45/36$

Then we have:

$h_{i} = h_{o} \cdot m$ $h_i=h_o*m$

$= 4 \cdot - \frac{45}{36}$ $=4*-45/36$

$= - \frac{180}{36}$ $=-180/36$

$= - 5$ $=-5$

So $h_{i} = - 5 cm$ $h_i=-5"cm"$

This means that the image is inverted.

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Question #c8fa6

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