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A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

Short Answer

Expert verified
The lens must move 6.30 mm toward the film.

Step by step solution

01

Identify Given Information

We are given that the focal length of the camera's lens is 90 mm. The object distance for the initial focus is 1.30 m, which is equivalent to 1300 mm, and the new object distance for refocusing is 6.50 m, which equals 6500 mm. We need to determine the change in image distance.
02

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, \( d_o \) is the object distance, and \( d_i \) is the image distance. We will use this formula twice: once for the initial object and once for the new object.
03

Calculate Initial Image Distance

Substitute the initial values into the lens formula: \( \frac{1}{90} = \frac{1}{1300} + \frac{1}{d_{i1}} \). Solving for \( d_{i1} \), we find that \( \frac{1}{d_{i1}} = \frac{1}{90} - \frac{1}{1300} \). Simplifying gives \( d_{i1} \approx 97.03 \) mm.
04

Calculate New Image Distance

Substitute the new object distance into the lens formula: \( \frac{1}{90} = \frac{1}{6500} + \frac{1}{d_{i2}} \). Solving for \( d_{i2} \), we find that \( \frac{1}{d_{i2}} = \frac{1}{90} - \frac{1}{6500} \). Simplifying gives \( d_{i2} \approx 90.73 \) mm.
05

Determine Change in Image Distance

Calculate the change in distance by subtracting the new image distance from the initial image distance: \( \Delta d_i = d_{i1} - d_{i2} = 97.03 - 90.73 = 6.30 \) mm. This indicates the lens must be moved 6.30 mm.
06

Direction of Lens Movement

Since the camera is focusing on a more distant object and the required image distance decreased, the lens must be moved closer to the film.

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Key Concepts

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

Lens Formula
In optics, the lens formula is a crucial equation that helps us relate key parameters of a lens-based system. These parameters include the focal length (\( f \)), object distance (\( d_o \)), and image distance (\( d_i \)). The formula is given by:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
This equation is derived from the geometry of lenses and the way they bend light. It essentially reflects how the curvature and position of an object impact where its image will form relative to the lens.
By rearranging this formula, you can solve for any one of these three variables if the other two are known. This makes it a fundamental tool in solving various optics problems, such as determining how much you need to move a camera lens when focusing on objects at different distances.
Focal Length
The focal length of a lens is a measure of how strongly the lens converges or diverges light. It is the distance from the lens to the point where parallel rays of light converge to a single point (focus). In our exercise with the camera, the focal length given is 90 mm.
Shorter focal lengths mean the lens has a stronger ability to bend light, which is useful for creating wider fields of view.
  • Positive focal length indicates a converging lens, which brings light to a point.
  • Negative focal length is typical for diverging lenses, spreading light apart.
Focal length is key in determining how far away you must be from a subject to bring it into clear focus. It also influences the size of the image formed on the film or sensor and is crucial for calculating both image and object distances using the lens formula.
Image Distance
Image distance, denoted as \( d_i \), is the distance from the lens to the point where the image is formed. This is a critical factor in determining the clarity and scale of the image captured by a camera.
Using the lens formula, we calculate image distances for both the initial and new focus positions in our problem. Initially with an object distance of 1.30 m, the image distance calculated was approximately 97.03 mm. With a change in the object distance to 6.50 m, the image distance shifted to 90.73 mm.
The change in image distance (\( \Delta d_i \)) can quantitatively show how the lens should be relocated within a camera apparatus. In this case, we determined the lens would need to move 6.30 mm closer to maintain focus on the object.
Object Distance
Object distance, usually symbolized as \( d_o \), is the essential distance between an object and a lens. This distance impacts where the image will form relative to the lens and thus is central to adjusting the focus.
  • In the initial scenario, our object was at 1.30 m (or 1300 mm) from the lens.
  • Refocusing was required for a new object distance of 6.50 m (or 6500 mm).
Changes in object distance directly affect the image distance, as evident from the change in focus involving moving the lens. Grasping this relationship allows photographers and optical engineers to manipulate lenses suitably for sharp imaging, efficiently using the lens formula to adjust focus settings. Understanding this interplay is central to mastering optics and ensuring clear, focused images.

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Most popular questions from this chapter

A converging lens with a focal length of 90.0 \(\mathrm{cm}\) forms an image of a 3.20 -cm-tall real object that is to the left of the lens. The image is 4.50 \(\mathrm{cm}\) tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the \(95.0-\mathrm{cm}\) lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?

A double-convex thin lens has surfaces with equal radii of curvature of magnitude \(2.50 \mathrm{cm} .\) Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.87 \(\mathrm{cm}\) from the lens. What is the index of refraction of the lens?

The focal length of the eyepiece of a certain microscope is 18.0 \(\mathrm{mm}\) . The focal length of the objective is 8.00 \(\mathrm{mm}\) . The distance between objective and eyepiece is \(19.7 \mathrm{cm} .\) The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

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