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Chapter 26: Problem 57

A converging lens has a focal length of \(88.00 \mathrm{cm} .\) An object \(13.0 \mathrm{cm}\) tall is located \(155.0 \mathrm{cm}\) in front of this lens. (a) What is the image distance? (b) Is the image real or virtual? (c) What is the image height? Be sure to include the proper algebraic sign.

Short Answer

Expert verified
(a) Image distance is 88.00 cm. (b) The image is real. (c) Image height is -7.38 cm (inverted).

Step by step solution

01

Use the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length of the lens, \( d_o \) is the object distance, and \( d_i \) is the image distance. We know \( f = 88.00 \text{ cm} \) and \( d_o = 155.0 \text{ cm} \). Substitute these values into the formula to find \( d_i \).
02

Solve for Image Distance \( d_i \)

Rearrange the formula to solve for \( d_i \): \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]Substitute \( f = 88.00 \text{ cm} \) and \( d_o = 155.0 \text{ cm} \) to find \( d_i \):\[ \frac{1}{d_i} = \frac{1}{88.00} - \frac{1}{155.0} \approx 0.01136 \]Inverting gives:\[ d_i \approx \frac{1}{0.01136} \approx 88.00 \text{ cm} \]
03

Determine Real or Virtual Image

Since \( d_i \) is positive (88.00 cm), the image is real and located on the opposite side of the lens from the object.
04

Calculate the Image Height

The magnification \( m \) of the lens is given by \( m = -\frac{d_i}{d_o} \). The image height \( h_i \) can be calculated as \( h_i = m \times h_o \), where \( h_o = 13.0 \text{ cm} \). First, calculate magnification:\[ m = -\frac{88.00}{155.0} \approx -0.5677 \]Now calculate the image height:\[ h_i = -0.5677 \times 13.0 \approx -7.38 \text{ cm} \]
05

Sign Convention and Conclusion

The negative sign in \( h_i \approx -7.38 \text{ cm} \) indicates that the image is inverted relative to the object. Thus, the height of the image is 7.38 cm, and it is inverted.

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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 crucial when working with lenses, especially converging lenses. It is mathematically expressed as \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here,
  • \( f \) is the focal length of the lens.
  • \( d_o \) is the distance from the object to the lens.
  • \( d_i \) is the distance from the lens to the image formed.
This formula allows us to determine any one of these three variables if the other two are known. To find the image distance \( d_i \), you can rearrange the formula:\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\]Substitute the known values to solve for the image distance.
Image Distance
In the exercise, we were tasked with finding the image distance, \( d_i \). Using the lens formula, we rearranged to solve for \( d_i \):\[\frac{1}{d_i} = \frac{1}{88.00} - \frac{1}{155.0} \]This calculation provides a value of \( 0.01136 \). To find \( d_i \), invert this value:\[ \frac{1}{0.01136} = 88.00 \text{ cm} \]This tells us that the image is formed 88.00 cm from the lens. The positive sign indicates that the image is on the opposite side of the lens from the object.
Image Magnification
Image magnification, \( m \), describes how much larger or smaller the image is compared to the object. It is calculated using the formula\[ m = -\frac{d_i}{d_o} \]The negative sign indicates the inversion of the image. For our problem, substituting the values we found:\[ m = -\frac{88.00}{155.0} \approx -0.5677 \]This tells us that the image is approximately 0.5677 times the size of the object. Consequently, it is smaller and inverted. Image height can also be calculated as:\[ h_i = m \times h_o \]Where \( h_o \) is the original object's height. Therefore:\[ h_i = -0.5677 \times 13.0 \approx -7.38 \text{ cm} \]
Real vs Virtual Image
Understanding whether an image is real or virtual is essential. In this context:
  • A real image is formed when the light converges at the image location.
  • A virtual image occurs where light appears to come from; it cannot be projected on a screen.
In our example, the image distance \( d_i \) was positive, which indicates a real image. This means the image forms on the opposite side of the lens where the rays actually converge. Real images are inverted relative to the object, as evidenced by the negative magnification and image height calculated in the previous section.

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Most popular questions from this chapter

Two converging lenses \(\left(f_{1}=9.00 \mathrm{cm}\right.\) and \(\left.f_{2}=6.00 \mathrm{cm}\right)\) are separated by \(18.0 \mathrm{cm} .\) The lens on the left has the longer focal length. An object stands \(12.0 \mathrm{cm}\) to the left of the left-hand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, (d) is the final image upright or inverted and (e) is it larger or smaller?

An object is placed in front of a converging lens in such a position that the lens \((f=12.0 \mathrm{cm})\) creates a real image located \(21.0 \mathrm{cm}\) from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens \((f=16.0 \mathrm{cm}) .\) A new, real image is formed. What is the image distance of this new image?

The moon's diameter is \(3.48 \times 10^{6} \mathrm{m},\) and its mean distance from the earth is \(3.85 \times 10^{8} \mathrm{m}\). The moon is being photographed by a camera whose lens has a focal length of \(50.0 \mathrm{mm}\). (a) Find the diameter of the moon's image on the slide film. (b) When the slide is projected onto a screen that is \(15.0 \mathrm{m}\) from the lens of the projector \((f=110.0 \mathrm{mm})\) what is the diameter of the moon's image on the screen?

An office copier uses a lens to place an image of a document onto a rotating drum. The copy is made from this image. (a) What kind of lens is used, converging or diverging? If the document and its copy are to have the same size, but are inverted with respect to one another, (b) how far from the document is the lens located and (c) how far from the lens is the image located? Express your answers in terms of the focal length \(f\) of the lens.

A tourist takes a picture of a mountain \(14 \mathrm{km}\) away using a camera that has a lens with a focal length of \(50 \mathrm{mm}\). She then takes a second picture when she is only \(5.0 \mathrm{km}\) away. What is the ratio of the height of the mountain's image on the camera's image sensor for the second picture to its height on the image sensor for the first picture?
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