# An object is placed in front of a concave mirror, 15.0 cm from the mirror's focal point. The...

## Question:

An object is placed in front of a concave mirror, 15.0 cm from the mirror's focal point. The image formed by the mirror is two times farther away from the focal point.

Calculate the focal length of the mirror There are actually two possible image distances di that satisfy the statement of the problem. Enter the larger of the two image distances. Enter the smaller of the two image distances. Show all work.

## Mirror equation:

For mirrors, the image distance and the object distance can be related to the focal length of the mirror by the equation:

{eq}\displaystyle \frac{1}{f} = \frac{1}{x} + \frac{1}{y} {/eq}

where:

• f is the focal length of the mirror
• we let x be the object distance
• we let y be the image distance

This equation is exactly the form of the thin lens equation as well. Sign convention is very important in the mirror equation. The sign of the focal length will determine whether the lens is concave (positive) or convex (negative). The sign of the image distance will tell if it is real (positive) or if it is virtual (negative).

Let us first work out the sign conventions that we will be using for the problem:

• If the object is in front of the mirror, then the object distance...

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