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Chapter 18: Problem 51

Calculate the image position and height. A \(3.0-\mathrm{cm}\) -tall object is \(45 \mathrm{cm}\) in front of a diverging mirror that has a \(-25 \mathrm{cm}\) focal length.

Short Answer

Expert verified
The image is found to be 37.5 cm in front of the mirror (virtual image) and it is 2.5 cm tall (upright)

Step by step solution

01

Identify given quantities

The focal length \(f\) is given as -25 cm while the object distance \(u\) is provided as 45 cm. The object height \(h_1\) is 3 cm. The negative signs indicate the object and the focus are in front of the mirror.
02

Calculate the image position

Applying the mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\), plug in \(-25\) as \(f\) and \(45\) as \(u\) and solve for \(v\), the distance of the image. The equation becomes \(\frac{1}{v} = \frac{1}{-25} - \frac{1}{45}\). Solving this equation yields \(v = -37.5 \, cm\) which means that the image is located 37.5 cm in front of the mirror.
03

Calculate image height

Utilize the magnification formula \(m = -h_2/h_1 = -v/u\). Here, \(h_2\) is the image height. From the formula above, we substitute \(v = -37.5cm\) and \(u = 45cm\), then we solve for \(h_2\). The equation becomes \(h_2 = m * h_1 = \frac{(- (-37.5 \, cm))}{45 \, cm} * 3.0 \, cm = 2.5 \, cm\). Therefore, the image height is +2.5cm. The positive sign means that the image will be upright.
04

Interpret the result

Based on these calculations, the image position and height are determined. The image is located 37.5 cm in front of the mirror which indicates that the mirror has produced a virtual image. The image will be upright and smaller than the object, with a height of 2.5 cm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Image Position
In the study of diverging mirrors, determining the image position is essential. The image position tells you where the image of an object appears. For diverging mirrors, this position is obtained using the mirror formula. The result is often a negative value, indicating that the image forms on the same side as the object. In our example, calculations using the mirror formula showed that the image position is -37.5 cm, meaning the image appears 37.5 cm in front of the mirror. This distance provides essential information on the nature of the image produced.
Mirror Formula
The mirror formula is a fundamental equation in optics, especially useful for dealing with diverging mirrors. It is represented as:\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]where:
  • \(f\) is the focal length of the mirror, which is negative for diverging mirrors.
  • \(v\) is the image distance from the mirror.
  • \(u\) is the object distance from the mirror.
This equation allows you to find unknown quantities like the image position, given other known values. For a diverging mirror, solving this equation typically results in a negative value for \(v\), indicating a virtual image.
Image Height
The height of an image formed by a mirror can tell you much about the nature of the image. The magnification formula links image height and object height:\[m = \frac{-h_2}{h_1} = \frac{-v}{u}\]Here:
  • \(h_2\) is the image height.
  • \(h_1\) is the object height.
  • \(m\) is magnification, which can be positive or negative.
In the example, with an object height of 3.0 cm and a calculated magnification of approximately 0.83, we found an image height of 2.5 cm. The positive height indicates that the image is upright.
Virtual Image
A virtual image is one that cannot be projected onto a screen since it appears to be on the opposite side of the mirror from the object. Diverging mirrors typically produce virtual images. These images appear upright and are located at the same side as the object. In our exercise, the image formed 37.5 cm in front of the mirror is a virtual image, a characteristic result of the negative image position obtained in the calculations.
Upright Image
In optics, an upright image means the image is oriented in the same direction as the object. For diverging mirrors, upright images are a common occurrence when the object is placed within the mirror's focal length. When performing calculations, a positive image height further confirms that the image is upright. In this scenario, the object height of 2.5 cm confirms its upright nature. This characteristic is valuable for applications where maintaining the orientation of the image is crucial.

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Most popular questions from this chapter

Calculate the image position and height. A 3.0 -cm-tall object is \(15 \mathrm{cm}\) in front of a converging mirror that has a \(25 \mathrm{cm}\) focal length.

You are using a converging lens to look at a splinter in your finger. The lens has a \(9.0 \mathrm{cm}\) focal length, and you place the splinter \(6.0 \mathrm{cm}\) from the lens. How far from the lens is the image? What is the magnification?

A lightbulb is \(60 \mathrm{cm}\) from a converging mirror with a focal length of \(20 \mathrm{cm} .\) Use ray tracing to determine the location of its image. Is the image upright or inverted? Is it real or virtual?

There is an interesting optical effect you have likely noticed while driving along a flat stretch of road on a sunny day. A small, distant dip in the road appears to be filled with water. You may even see the reflection of an oncoming car. But, as you get closer, you find no puddle of water after all; the shimmering surface vanishes, and you see nothing but empty road. It was only a mirage, the name for this phenomenon. The mirage is due to the different index of refraction of hot and cool air. The actual bending of the light rays that produces the mirage is subtle, but we can make a simple model as follows. When air is heated, its density decreases and so does its index of refraction. Consequently, a pocket of hot air in a dip in a road has a lower index of refraction than the cooler air above it. Incident light rays with large angles of incidence (that is, nearly parallel to the road, as shown in Figure P18.83) experience total internal reflection. The mirage that you see is due to this reflection. As you get nearer, the angle goes below the critical angle and there is no more total internal reflection; the 鈥渨ater鈥� disappears! Which of these changes would allow you to get closer to the mirage before it vanishes? A. Making the pocket of hot air nearer in temperature to the air above it B. Looking for the mirage on a windy day, which mixes the air layers C. Increasing the difference in temperature between the pocket of hot air and the air above it D. Looking at it from a greater height above the ground

A dentist uses a curved mirror to view the back side of teeth on the upper jaw. Suppose she wants an erect image with a magnification of 2.0 when the mirror is \(1.2 \mathrm{cm}\) from a tooth. (Treat this problem as though the object and image lie along a straight line.) Use ray tracing to decide whether a converging or diverging mirror is needed, and to estimate its focal length.
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