# 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?

Apr 30, 2014

+15.0 cm.

#### Explanation:

For this question, we need to use the mirror formula

• $\setminus \frac{1}{f} = \frac{1}{d} _ o + \frac{1}{d} _ i$.

What the problem gives us is: f = 10.0 cm, and ${d}_{i}$ = +30.0 cm. We know that ${d}_{i}$ is positive because it forms a real image. So we are solving for ${d}_{o}$. Isolating the unknown to its own side of the equation, in this case by subtracting $\frac{1}{d} _ i$ from both sides, will accomplish this.

• $\setminus \frac{1}{{d}_{o}} = \setminus \frac{1}{f} - \frac{1}{d} _ i$

• $\frac{1}{d} _ o = \frac{1}{10} - \frac{1}{30}$. FInd a common denominator.

• $\frac{1}{d} _ o = \frac{3}{30} - \frac{1}{30}$

• $\frac{1}{d} _ o = \frac{2}{30}$. To find ${d}_{o}$, take the reciprocal.

• ${d}_{o} = \frac{30}{2} = + 15.0 c m$

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