If an object is placed 25.0 cm from a concave mirror whose focal length is 5.0 cm, where is the image located?

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If an object is placed 25.0 cm from a concave mirror whose focal length is 5.0 cm, where is the image located?

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Answer 1 · 2014-04-30T21:38:05

For this question, we need to use the mirror formula $\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d_{i}}$ $1/f= 1/d_o +1/d_i$ .

What the problem gives us is: f = 5.0 cm, and $d_{o}$ $d_o$ = 25.0 cm. So we are solving for $d_{i}$ $d_i$ . Isolating the unknown to its own side of the equation, in this case by subtracting $\frac{1}{d_{i}}$ $1/d_i$ from both sides, will accomplish this.

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

$\frac{1}{d_{i}} = \frac{1}{5} - \frac{1}{25}$ $1/d_i = 1/5 − 1/25$ . FInd a common denominator.

$\frac{1}{d_{i}} = \frac{5}{25} - \frac{1}{25}$ $1/d_i = 5/25 − 1/25$

$\frac{1}{d_{i}} = \frac{4}{25}$ $1/d_i = 4/25$ . To find $d_{i}$ $d_i$ , take the reciprocal.

$d_{i} = \frac{25}{4} = + 6.25 c m$ $d_i = 25/4 = +6.25 cm$

We know that this is a real image because $d_{i}$ $d_i$ is positive.

The same process can be used if you know the distance from the image to the vertex of the mirror, and are looking for $d_{o}$ $d_o$ .

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If an object is placed 25.0 cm from a concave mirror whose focal length is 5.0 cm, where is the image located?

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