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Chapter 31: Problem 44

An object \(15 \mathrm{cm}\) from a concave mirror has a virtual image magnified 2.5 times. What's the mirror's focal length?

Short Answer

Expert verified
The focal length of the mirror is \(-18.75 cm\).

Step by step solution

01

Mirror and Magnification Formulas

First of all, do know the mirror formula: \[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\], where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. Also, the magnification formula is: \[m = -\frac{v}{u}\]. The magnification given is positive with the mirror described as 'concave', which indicates a virtual, upright image. This gives us \(u = -15 cm\), and \(m = 2.5\).
02

Calculate the image distance

Use the magnification formula to find the image distance \(v\). Substituting \(-2.5\) (negative as the object is real) for \(m\), and \(-15 cm\) for \(u\) in the magnification formula yields \(v = -2.5 \times -15 cm\), so \(v = 37.5 cm\). Remember, in mirror formulae, the sign conventions are: object distance is always negative, and for real images, image distance is negative. But since our image is virtual, \(v = 37.5 cm\) is positive.
03

Calculating the focal length

Now substituting \(v = 37.5 cm\) and \(u = -15 cm\) in the mirror formula yields: \[\frac{1}{f} = \frac{1}{37.5} + \frac{1}{-15}\]. Solving this gives us \(f = -18.75 cm\) (negative as it is a concave mirror).

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