/*! 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 60 When a converging lens is used i... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 60

When a converging lens is used in a camera (as in Figure \(26.26 b\) ), the film must be at a distance of \(0.210 \mathrm{m}\) from the lens to record an image of an object that is \(4.00 \mathrm{m}\) from the lens. The same lens and film are used in a projector (see Figure \(26.27 b\) ), with the screen \(0.500 \mathrm{m}\) from the lens. How far from the projector lens should the film be placed?

Short Answer

Expert verified
The film should be placed 0.333 m from the projector lens.

Step by step solution

01

Understand the Lens Formula

For both camera and projector scenarios, the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) is used, where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance.
02

Find Focal Length for Camera

Given \(d_o = 4.00\, \text{m} \) and \(d_i = 0.210\, \text{m}\), apply the formula to find \( f \): \[ \frac{1}{f} = \frac{1}{4.00} + \frac{1}{0.210} \]Calculate \(f\).
03

Calculate Focal Length

Calculating further: \[ \frac{1}{f} = 0.25 + 4.76 \approx 5.01 \] Thus, \( f \approx 0.20\, \text{m} \).
04

Using Focal Length in Projector

For the projector, \(d_i = 0.500\, \text{m} \). Using \( f = 0.20\, \text{m} \), find \(d_o\):\[ \frac{1}{0.20} = \frac{1}{d_o} + \frac{1}{0.500} \].
05

Solve for Object Distance in Projector

Rearranging gives: \[ \frac{1}{d_o} = \frac{1}{0.20} - \frac{1}{0.500} \]Calculate to find \(d_o\).
06

Calculate Projector Object Distance

Compute \[ \frac{1}{d_o} = 5 - 2 = 3 \] Thus, \(d_o = \frac{1}{3} \approx 0.333\, \text{m} \).
07

Conclusion

The film should be placed \(0.333\, \text{m} \) away from the projector lens.

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Key Concepts

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

Focal Length
The focal length of a lens is a critical parameter that determines how light rays converging through the lens are focused. In essence, it is the distance from the lens at which parallel rays of light converge to a single point. This point is known as the focal point. The focal length (\( f \)) determines the power of the lens, with shorter focal lengths resulting in more powerful focusing capabilities.

For cameras and projectors, knowing the focal length helps in predicting and manipulating how images form on film or other surfaces.
  • The formula for the focal length is given by the lens maker's equation, often utilized as: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance, and \( d_i \) is the image distance.
  • In our example, the focal length was calculated to be \( 0.20 \, \text{m} \), which was determined based on the object (\( 4.00 \, \text{m} \)) and image distances (\( 0.210 \, \text{m} \)) of the camera setup.
Understanding this allows one to design systems where images can be crisp and appropriately scaled for visualization.
Converging Lens
A converging lens, often referred to as a convex lens, is specially shaped to focus incoming parallel light rays into a single point known as the focal point. This effect happens because the lens is thicker in the middle than at the edges. Such lenses are crucial in both cameras and projectors. Without them, creating a sharp image on film or screen would be impossible.

Whether you're snapping photos or projecting a movie onto a screen, converging lenses help shape and focus the light so that images can form.
  • Converging lenses bring several practical advantages, including the ability to form real and inverted images at specific distances.
  • In the given exercise, a converging lens ensures the film in the camera correctly captures the image, and in a projector, it ensures the image falls neatly on the screen at the appropriate distance.
The ability of the lens to magnify an object makes it indispensable in optical devices.
Image Distance
Image distance (\( d_i \)) is a pivotal concept in optics that tells us how far the image is formed from the lens. This distance is crucial because it determines where the image will be in focus, whether on camera film or a projection screen.

In cameras, the image distance dictates the positioning of the film to ensure the image appears sharp and focused. Similarly, in projectors, it determines where the screen should be placed.
  • From the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), once the focal length and object distance are known, you can solve for \( d_i \).
  • In our exercise example, the image distance for the camera was \( 0.210 \, \text{m} \) while for the projector, a calculated adjustment to \( 0.333 \, \text{m} \) ensured a focused presentation.
Mastering image distance is key to achieving clarity in any optical setup.

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Most popular questions from this chapter

Two converging lenses are separated by \(24.00 \mathrm{cm} .\) The focal length of each lens is \(12.00 \mathrm{cm} .\) An object is placed \(36.00 \mathrm{cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

A ray of light impinges from air onto a block of ice \((n=1.309)\) at a \(60.0^{\circ}\) angle of incidence. Assuming that this angle remains the same, find the difference \(\theta_{2, \text { ice }}-\theta_{2, \text { water }}\) in the angles of refraction when the ice turns to water \((n=1.333)\)

An object is \(18 \mathrm{cm}\) in front of a diverging lens that has a focal length of \(-12 \mathrm{cm} .\) How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of \(2.0 ?\)

The distance between the lenses in a compound microscope is \(18 \mathrm{cm} .\) The focal length of the objective is \(1.5 \mathrm{cm} .\) If the microscope is to provide an angular magnification of -83 when used by a person with a normal near point \((25 \mathrm{cm}\) from the eye), what must be the focal length of the eyepiece?

An object is placed \(20.0 \mathrm{cm}\) to the left of a diverging lens \((f=\) \(-8.00 \mathrm{cm}$$) .\) A concave mirror \((f=12.0 \mathrm{cm})\) is placed \(30.0 \mathrm{cm}\) to the right of the lens. (a) Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?
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