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To focus a camera on objects at different distances, the converging lens is moved toward or away from the image sensor, so a sharp image always falls on the sensor. A camera with a telephoto lens \((f=200.0 \mathrm{mm})\) is to be focused on an object located first at a distance of \(3.5 \mathrm{m}\) and then at \(50.0 \mathrm{m}\). Over what distance must the lens be movable?

Short Answer

Expert verified
The lens must move 11 mm.

Step by step solution

01

Understand the Lens Formula

To solve this problem, we need to use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length of the lens, \( d_o \) is the object distance, and \( d_i \) is the image distance.
02

Calculate Image Distance for First Object Location

Substitute \( f = 200.0 \mathrm{mm} \) and \( d_o = 3.5 \mathrm{m} = 3500 \mathrm{mm} \) into the lens formula. Solve for \( d_i \):\[ \frac{1}{200} = \frac{1}{3500} + \frac{1}{d_i} \]Simplify and solve to find \( d_i \).
03

Solve for Image Distance for \( d_o = 3.5m \)

Rearrange the formula:\[ \frac{1}{d_i} = \frac{1}{200} - \frac{1}{3500} \]Calculate \( \frac{1}{d_i} = 0.005 - 0.0002857 \approx 0.0047143 \). Thus, \( d_i = \frac{1}{0.0047143} \approx 212 \mathrm{mm} \).
04

Calculate Image Distance for Second Object Location

Repeat the process for the second object location at \( d_o = 50.0 \mathrm{m} = 50000 \mathrm{mm} \).\[ \frac{1}{200} = \frac{1}{50000} + \frac{1}{d_i} \]Solve this for \( d_i \).
05

Solve for Image Distance for \( d_o = 50.0m \)

Rearrange the formula:\[ \frac{1}{d_i} = \frac{1}{200} - \frac{1}{50000} \]Calculate \( \frac{1}{d_i} = 0.005 - 0.00002 \approx 0.00498 \). Thus, \( d_i = \frac{1}{0.00498} \approx 201 \mathrm{mm} \).
06

Determine the Distance the Lens Must Move

To find how much the lens must be moved, subtract the two computed image distances:\( 212 \mathrm{mm} - 201 \mathrm{mm} = 11 \mathrm{mm} \)This is the distance the lens must be able to move to focus on objects at both given distances.

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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 crucial concept in optics, as it defines how strongly the lens converges or diverges light. In simpler terms, it determines the distance over which parallel rays of light are brought into focus. A shorter focal length implies that the lens has more power to bend light quickly, bringing it to a focus over a shorter distance. Conversely, a longer focal length means the lens bends light less sharply, requiring a longer distance.
  • The focal length is usually denoted by the letter \( f \).
  • It can be expressed in different units, such as millimeters (as in cameras) or meters (as more commonly used in certain optics applications).
  • In the lens formula, the focal length is a fixed characteristic of the lens.
For practical applications, understanding focal length helps in selecting the right lens for the desired camera setup or other optical devices.
Image Distance
Image distance refers to the distance from the lens to the image sensor or film where the image is formed. In terms of optics, this concept helps determine the positioning of the image for clarity and focus. This distance is represented by \( d_i \) in the lens formula.
  • Image distance will vary based on the object distance and the lens's focal length.
  • It is measured from the lens along the direction in which the image is formed.
When using the lens formula, a calculated image distance tells you how far the lens must be from the film or sensor to capture a sharp image. If you want a clearer understanding, imagine adjusting a microscope or binoculars: the crispness of what you see depends on this critical measurement.
Object Distance
Object distance, denoted by \( d_o \), represents how far an object is located from the lens. In optical setups like cameras or telescopes, this measurement is pivotal in determining how focused an object will appear when viewed through the lens.
  • The object distance directly impacts image distance and how light is refracted through the lens.
  • An object's distance can be physically altered to achieve different imaging results without changing the lens.
In practical terms, it means moving an object closer or further away from the lens to bring it into focus. Think of it as adjusting your position in relation to your smartphone camera to take a more detailed picture. This movement changes the object distance, which the lens formula can use to calculate the correct settings for getting a sharp image.
Optics
Optics is the branch of physics that studies light and its interactions with matter. It encompasses a range of principles like reflection, refraction, and diffraction, all of which are vital for designing optical instruments such as glasses, cameras, and telescopes. The lens formula ties into optics, providing the framework for understanding how lenses manipulate light to form clear images. The essential elements of the lens formula in optics include:
  • Refraction: The bending of light as it passes through different media. This is the basis for how lenses focus light.
  • Lenses: An essential component in optics that focuses or disperses light. They can be convex or concave, affecting their focusing properties.
  • Image Formation: How lenses create an image on the sensor or film, requiring precise measurement of object and image distance.
In your everyday life, optics comes into play with every pair of eyeglasses, the lens in a camera, and the telescopes used by astronomers to gaze at the stars. Understanding these principles not only helps in solving textbook exercises but also enhances your grasp of the technology around us.

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Most popular questions from this chapter

Red light \((n=1.520)\) and violet light \((n=1.538)\) traveling in air are incident on a slab of crown glass. Both colors enter the glass at the same angle of refraction. The red light has an angle of incidence of \(30.00^{\circ} .\) What is the angle of incidence of the violet light?

The back wall of a home aquarium is a mirror that is a distance of \(40.0 \mathrm{cm}\) away from the front wall. The walls of the tank are negligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. The index of refraction of air is \(n_{\text {air }}=1.000\) and that of water is \(n_{\text {water }}=1.333\). (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall.

A narrow beam of light from a laser travels through air \((n=1.00)\) and strikes point A on the surface of the water \((n=1.33)\) in a lake. The angle of incidence is \(55^{\circ} .\) The depth of the lake is \(3.0 \mathrm{m} .\) On the flat lake-bottom is point B, directly below point A. (a) If refraction did not occur, how far away from point \(B\) would the laser beam strike the lake- bottom? (b) Considering refraction, how far away from point B would the laser beam strike the lake-bottom?

A layer of oil \((n=1.45)\) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid. The angle of incidence is \(64.0^{\circ},\) and the angle of refraction is \(53.0^{\circ} .\) What is the index of refraction of the unknown liquid?

An amateur astronomer decides to build a telescope from a discarded pair of eyeglasses. One of the lenses has a refractive power of 11 diopters, and the other has a refractive power of 1.3 diopters. (a) Which lens should be the objective? (b) How far apart should the lenses be separated? (c) What is the angular magnification of the telescope?

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