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Chapter 38: Problem 34

Describe fully the image of an object that is \(10 \mathrm{~cm}\) high and 28 \(\mathrm{cm}\) from a diverging lens of focal length \(-7.0 \mathrm{~cm}\).

Short Answer

Expert verified
The image is 2 cm tall, virtual, erect, and 5.6 cm from the lens.

Step by step solution

01

Understand the Problem

We have an object that is 10 cm tall and located 28 cm from a diverging lens. The focal length of the lens is -7.0 cm. We need to find the characteristics (position, size, orientation, and type) of the image formed by the lens.
02

Use the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Here, \( f = -7.0 \) cm and \( u = -28 \) cm (we take the object distance as negative since the object is in front of the lens). Substitute the known values to find \( v \):\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-7} + \frac{1}{-28} \]
03

Simplify and Solve for the Image Distance

Calculate \( \frac{1}{v} \):\[ \frac{1}{v} = -\frac{1}{7} - \frac{1}{28} = -\frac{4}{28} - \frac{1}{28} = -\frac{5}{28} \]Thus, \( v = -\frac{28}{5} = -5.6 \) cm. The negative sign indicates that the image is virtual and on the same side of the lens as the object.
04

Determine the Magnification

The magnification \( m \) is given by \( m = \frac{v}{u} \). Substitute \( v = -5.6 \) cm and \( u = -28 \) cm:\[ m = \frac{-5.6}{-28} = \frac{5.6}{28} = 0.2 \]
05

Calculate the Image Height

The image height \( h' \) can be found using the magnification formula, \( m = \frac{h'}{h} \), where \( h \) is the object height. With \( m = 0.2 \) and \( h = 10 \) cm:\[ h' = m \times h = 0.2 \times 10 = 2 \text{ cm} \]
06

Analyze the Results

The image is 2 cm tall, virtual (since the image distance is negative), erect (since the magnification is positive), and located 5.6 cm on the same side of the lens as the object.

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Key Concepts

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

Lens Formula
The lens formula is a crucial part of understanding how lenses form images. It is represented by the equation:
  • \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
where:
  • \( f \) is the focal length of the lens, which indicates how strongly the lens can converge or diverge light.
  • \( v \) is the image distance, meaning the distance from the lens to the image formed.
  • \( u \) is the object distance, which is how far the object is from the lens.
For a diverging lens, the focal length \( f \) is negative. This formula helps us determine the position and nature of the image formed by a lens when knowing the object distance and the lens's focal length. It is an invaluable tool in geometrical optics for predicting and understanding image characteristics.
Diverging Lens
A diverging lens, also known as a concave lens, spreads out light rays that have been refracted through it. These lenses are thinner at the center than at the edges.
The key features of a diverging lens include:
  • It has a negative focal length. This means it cannot converge parallel light rays to a single point.
  • The images formed by a diverging lens are typically virtual, reduced in size, and erect.
Diverging lenses are typically used in eyeglasses for nearsighted individuals and also help correct certain image distortions in optical devices. Understanding the nature of diverging lenses is essential in predicting how they interact with light and form images.
Virtual Image
In optics, a virtual image is formed when the light rays diverge, and the image cannot be projected onto a screen. For diverging lenses, the image appears to be on the same side as the object. Characteristics of virtual images include:
  • They have a negative image distance \( v \), indicating that they are formed on the same side of the lens as the incoming light.
  • Virtual images are erect, meaning that the image has the same orientation as the object.
  • They are reduced in size compared to the actual object.
Virtual images, despite not being real projections, are vital in understanding how lenses manipulate light. These images are a common result when using diverging lenses and play a significant role in various optical instruments and vision correction technologies.

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Most popular questions from this chapter

A double convex glass lens \((n=1.50)\) has faces of radius \(8 \mathrm{~cm}\) each. Compute its focal length in air and when immersed in water \((n=1.33)\).

An object is very far from the front of a thin converging lens. As the object approaches the lens, still farther than one focal length from it, what happens to the size of the image?

A thin glass lens \((n=1.50)\) has a focal length of \(+10 \mathrm{~cm}\) in air. Compute its focal length in water \((n=1.33)\). Using $$\frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$ we get $$\begin{array}{ll} \text { For air: } & \frac{1}{10}=(1.50-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \text { For water: } & \frac{1}{f}=\left(\frac{1.50}{1.33}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \end{array}$$ Divide one equation by the other to obtain \(f=5.0 / 0.128=39 \mathrm{~cm}\).

Two thin lenses, of focal lengths \(+12\) and \(-30 \mathrm{~cm}\), are in contact. Compute the focal length and power of the combination.

A thin lens has a focal length of \(+20.0 \mathrm{~cm}\). An object is placed 300 \(\mathrm{m}\) in front of the lens. Roughly where will the image be formed? Explain your answer.
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