If a concave mirror with a focal length of 10.0 cm creates a real image 30.0 cm away on its principal axis, how far from the mirror is the corresponding object?

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If a concave mirror with a focal length of 10.0 cm creates a real image 30.0 cm away on its principal axis, how far from the mirror is the corresponding object?

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Answer 1 · 2014-04-30T20:20:13

+15.0 cm.

Explanation:

For this question, we need to use the mirror formula

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

What the problem gives us is: f = 10.0 cm, and $d_{i}$ $d_i$ = +30.0 cm. We know that $d_{i}$ $d_i$ is positive because it forms a real image. So we are solving for $d_{o}$ $d_o$ . 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.

$\frac{1}{d_{o}} = \frac{1}{f} - \frac{1}{d_{i}}$ $\frac{1}{d_o} = \frac{1}{f} - 1/d_i$
$\frac{1}{d_{o}} = \frac{1}{10} - \frac{1}{30}$ $1/d_o = 1/10 -1/30$ . FInd a common denominator.
$\frac{1}{d_{o}} = \frac{3}{30} - \frac{1}{30}$ $1/d_o = 3/30 - 1/30$
$\frac{1}{d_{o}} = \frac{2}{30}$ $1/d_o = 2/30$ . To find $d_{o}$ $d_o$ , take the reciprocal.
$d_{o} = \frac{30}{2} = + 15.0 c m$ $d_o = 30/2 = +15.0 cm$

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

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If a concave mirror with a focal length of 10.0 cm creates a real image 30.0 cm away on its principal axis, how far from the mirror is the corresponding object?

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