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Chapter 23: Problem 83

A dentist holds a small mirror \(1.9 \mathrm{cm}\) from a surface of a patient's tooth. The image formed is upright and 5.0 times as large as the object. (a) Is the image real or virtual? (b) What is the focal length of the mirror? Is it concave or convex? (c) If the mirror is moved closer to the tooth, does the image get larger or smaller? (d) For what range of object distances does the mirror produce an upright image?

Short Answer

Expert verified
Answer: The image formed by the mirror is virtual, the mirror's focal length is approximately 1.58 cm, and the mirror used is a convex mirror.

Step by step solution

01

Identify the known values and the desired outcomes.

The distance of the object from the mirror, \(o\), is given as \(1.9\,\mathrm{cm}\). The magnification, \(m\), of the image is given as \(5.0\) times the object size, which means it's an enlarged image. We need to find the image's type, the mirror's focal length, the mirror's type, and the effect of changing the object distance on the image size.
02

Determine the nature of the image.

When the image formed by a mirror is upright, it is always a virtual image. Therefore, the image in this case is virtual.
03

Determine the distance of the image from the mirror, i.

We are given the magnification, \(m\), and since the image is upright and enlarged, the magnification will be positive. Using the magnification formula, \(m = -\frac{i}{o}\), we can find the image's distance from the mirror: \(m = -\frac{i}{o} \Rightarrow i = -mo\) Plugging in the known values: \(i = -(5.0)(-1.9\,\mathrm{cm})\) \(i = 9.5\,\mathrm{cm}\) The image is \(9.5\,\mathrm{cm}\) from the mirror.
04

Determine the focal length, f, of the mirror.

Applying the mirror formula, \(\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\), we can find the focal length, \(f\), of the mirror: \(\frac{1}{f} = \frac{1}{1.9\,\mathrm{cm}} + \frac{1}{9.5\,\mathrm{cm}}\) To find the value of \(f\), we will first find a common denominator for the fractions, which will be \(1.9 \cdot 9.5\): \(\frac{1}{f} = \frac{9.5 + 1.9}{1.9\,\mathrm{cm} \cdot 9.5\,\mathrm{cm}}\) \(\frac{1}{f} = \frac{11.4}{18.05}\,\,\mathrm{cm}^{-1}\) Now, we will find \(f\) by taking the reciprocal: \(f = \frac{18.05}{11.4}\,\mathrm{cm}\) \(f \approx 1.58\,\mathrm{cm}\) The focal length of the mirror is approximately \(1.58\,\mathrm{cm}\).
05

Determine the type of mirror.

Since the image formed is virtual and upright, the mirror must be a convex mirror. Concave mirrors produce real and inverted images.
06

Analyze the effect of moving the mirror closer to the tooth.

If the mirror is moved closer to the tooth, the value of \(o\) will decrease. For a convex mirror, decreasing the distance of the object from the mirror will cause the image to get smaller.
07

Determine the range of object distances that produce an upright image.

Since a convex mirror always produces an upright and virtual image, the range of object distances that produce an upright image is any distance from the mirror. In summary, the image formed is virtual and the mirror used is a convex mirror with a focal length of approximately \(1.58\,\mathrm{cm}\). When the mirror is moved closer to the tooth, the image size will get smaller. Any object distance will produce an upright image when a convex mirror is used.

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