/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A diverging lens has a focal len... [FREE SOLUTION] | 91影视

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Chapter 23: Problem 33

A diverging lens has a focal length of \(20.0 \mathrm{~cm}\). Locate the images for object distances of (a) \(40.0 \mathrm{~cm}\), (b) \(20.0 \mathrm{~cm}\), and (c) \(10.0 \mathrm{~cm} .\) For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.

Short Answer

Expert verified
For all three cases, (a), (b), and (c), the images formed by the diverging lens are virtual and upright. The image distances are \(-13.33cm\), \(-10cm\), and \(-6.67cm\) respectively, and the magnifications are -0.333, -0.5, and -0.667 respectively.

Step by step solution

01

Problem (a) Solution

The object distance, \(u\) is \(-40.0cm\) (negative because the object is on the same side as incoming light before refraction). Using the lens formula, \(1/f=1/v-1/u\), plug \(f=-20cm\) (\(f\) is negative for diverging lenses), and \(u=-40cm\) to get the image distance \(v\) which equals \(-13.33cm\). This is virtual because \(v\) is negative. Finally, compute the magnification, \(m\), using the formula \(-v/u\) to get -0.333. The negative sign indicates the image is upright.
02

Problem (b) Solution

The object distance, \(u\) is \(-20.0cm\). Substituting \(f=-20cm\) and \(u=-20cm\) into the lens formula gives the image distance, \(v=-10cm\). Again this is a virtual image since \(v\) is negative. Magnification, \(m=-v/u=-0.5\), so the image is upright.
03

Problem (c) Solution

The object distance, \(u\) is \(-10.0cm\). Inserting \(f=-20cm\) and \(u=-10cm\) in the lens formula gives the image distance, \(v=-6.67cm\). Since \(v\) is negative, the image is virtual. Magnification, \(m=-v/u=-0.667\), shows the image is upright.

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Key Concepts

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

Understanding Focal Length
The focal length of a lens is a fundamental property that determines how the lens will bend light. For a diverging lens, this is an imaginary point from which light appears to be diverging. This is why it is given as a negative value. In the exercise, the diverging lens has a focal length of \(-20.0 \, \text{cm}\). Here, the negative sign indicates divergence, meaning that the lens spreads out light rays.

It's important to remember:
  • Diverging lenses have negative focal lengths.
  • The shorter the focal length, the stronger the lens鈥檚 power to spread light.
Understanding this concept is crucial in predicting how the light will behave as it passes through the lens.
What is a Virtual Image?
A virtual image appears to be located inside the lens or on the same side as the object, which is different from a real image that can be projected onto a screen. In this exercise, all images produced by the diverging lens are virtual.

This happens because diverging lenses always spread light rays. Consequently, the light rays never intersect on the opposite side of the lens. Instead, they seem to meet when traced backwards, forming a virtual image.

Key points:
  • Virtual images cannot be captured on a screen.
  • Diverging lenses always produce virtual images and are upright.
  • These images appear smaller than the actual object, depending on their distance from the lens.
Exploring Magnification
Magnification tells us how much larger or smaller the image is compared to the object. It is expressed by the ratio of the image height to the object height, or alternatively, the ratio of the image distance \(v\) to the object distance \(u\).

In this problem:
  • Magnification was calculated using \(m = -\frac{v}{u}\).
  • A negative magnification indicates the image is upright.
  • The absolute value of magnification shows the size relative to the object.
For example, in part (a), with \(m = -0.333\), the image is upright and roughly one-third the size of the object. Understanding magnification helps in visualizing the size and orientation of the image relative to the object.

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Most popular questions from this chapter

An object is placed \(15.0 \mathrm{~cm}\) from a first converging lens of focal length \(10.0 \mathrm{~cm} .\) A second converging lens with focal length \(5.00 \mathrm{~cm}\) is placed \(10.0 \mathrm{~cm}\) to the right of the first converging lens. (a) Find the position \(q_{1}\) of the image formed by the first converging lens. (b) How far from the second lens is the image of the first lens? (c) What is the value of \(p_{2}\), the object position for the second lens? (d) Find the position \(q_{2}\) of the image formed by the second lens. (e) Calculate the magnification of the first lens. (f) Calculate the magnification of the second lens. (g) What is the total magnification for the system? (h) Is the final image real or virtual? Is it upright or inverted?

Object \(\mathrm{O}_{1}\) is \(15.0 \mathrm{~cm}\) to the left of a converging lens with a \(10.0-\mathrm{cm}\) focal length. A second lens is positioned \(10.0 \mathrm{~cm}\) to the right of the first lens and is observed to form a final image at the position of the original object \(\mathrm{O}_{1}\). (a) What is the focal length of the second lens? (b) What is the overall magnification of this system? (c) What is the nature (i.e., real or virtual, upright or inverted) of the final image?

An object of height \(2.0 \mathrm{~cm}\) is placed \(30.0 \mathrm{~cm}\) from a convex mirror of focal length having magnitude \(10.0 \overline{\mathrm{cm}}\). (a) Find the location of the image. (b) Describe the properties of the image.

Two plane mirrors stand facing each other, \(3.0 \mathrm{~m}\) apart, and a woman stands between them. The woman faces one of the mirrors from a distance of \(1.0 \mathrm{~m}\), with the palm of her left hand facing the closer mirror. (a) What is the apparent position of the closest image of her left hand, measured from the surface of the mirror in front of her? Does it show the palm of her hand or the back of her hand? (b) What is the position of the next image? Does it show the palm of her hand or the back of her hand? (c) Repeat for the third image. (d) Which of the images are real and which are virtual?

A contact lens is made of plastic with an index of refraction of \(1.50 .\) The lens has an outer radius of curvature of \(+2.00 \mathrm{~cm}\) and an inner radius of curvature of \(+2.50 \mathrm{~cm}\). What is the focal length of the lens?
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