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Chapter 31: Problem 41

An object's image in a 27 -cm-focal-length concave mirror is upright and magnified by a factor of \(3 .\) Where is the object?

Short Answer

Expert verified
The object is approximately 13.5 cm in front of the mirror.

Step by step solution

01

Identify the given

In this case, we have a concave mirror with a focal length (f) of -27 cm (as it is concave, the focal length is considered as negative), and the magnification (m) is -3 (as the image is upright and magnified, the magnification is considered as negative). Our goal is to find the object distance (do) .
02

Apply the mirror equation

The mirror equation is given by \(1/f = 1/do + 1/di\) where f is the focal length, do is the object distance and di is the image distance. Our task is to find the object distance so we'll rearrange the mirror equation to find it as \(do = 1/(1/f - 1/di)\). We know the focal length but we don't know the image distance yet.
03

Apply the magnification formula

The magnification (m) in a mirror is the negative ratio of image distance to object distance. This gives us \(m = -di/do\). Since we know the magnification is -3, we can rearrange to find the image distance as \(di = -m * do\).
04

Substitute values

Now we substitute the image distance from Step 3 into the rearranged mirror equation from Step 2. That allows us to solve for the object distance using only known values and the object distance itself which leads us to this equation \(do =1/ (1/-27) - 1/((-3) * do))\).
05

Simplify and solve the equation

Solving this equation will give us the object's position. Since it appears to be a quadratic equation, we can solve for \(do\) by rearranging it to a standard quadratic form and then using the quadratic formula or factoring. This will lead us to find that the object distance \(do\) is approximately -13.5 cm.

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