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

Short Answer

Expert verified
The sensor position changes by -3.0 mm; the lens moves toward the sensor.

Step by step solution

01

Understanding the Lens Formula

The lens formula relates the object distance (\(d_o\)), the image distance (\(d_i\)), and the focal length (\(f\)) of the lens: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Given the focal length \(f = 90 \text{ mm} = 0.09 \text{ m}\). We first need to calculate \(d_i\) for both object distances.
02

Calculate Image Distance for First Object

Given the first object distance \(d_{o1} = 1.30 \text{ m}\), use the lens formula to find \(d_{i1}\).\[ \frac{1}{0.09} = \frac{1}{1.30} + \frac{1}{d_{i1}} \]Simplifying:\[ \frac{1}{d_{i1}} = \frac{1}{0.09} - \frac{1}{1.30} \approx 10.6528 \]\[ d_{i1} \approx 0.0939 \text{ m} \text{ (or 93.9 mm)} \]
03

Calculate Image Distance for Second Object

Now calculate \(d_{i2}\) for the second object distance \(d_{o2} = 6.50 \text{ m}\).\[ \frac{1}{0.09} = \frac{1}{6.50} + \frac{1}{d_{i2}} \]Simplifying:\[ \frac{1}{d_{i2}} = \frac{1}{0.09} - \frac{1}{6.50} \approx 11.0051 \]\[ d_{i2} \approx 0.0909 \text{ m} \text{ (or 90.9 mm)} \]
04

Determine Change in Sensor Position

The camera's sensor needs to move based on the change in image distance: \[ \Delta d_i = d_{i2} - d_{i1} = 90.9 \text{ mm} - 93.9 \text{ mm} = -3.0 \text{ mm} \]A negative value indicates that the lens must move toward the sensor.

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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 a fundamental concept in optics used to relate the characteristics of lenses, such as the object distance, image distance, and the lens's focal length. It's 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, and \(d_i\) is the distance from the lens to the image. This formula is essential in determining how lenses form images, whether in cameras, glasses, or microscopes.
  • The focal length \(f\) is crucial: a positive value indicates a converging lens, while a negative value indicates a diverging lens.
  • The object distance \(d_o\) is always considered positive if the object is in front of the lens.
  • The image distance \(d_i\) can be positive or negative, indicating real and virtual images, respectively.
By rearranging the lens formula, you can calculate the unknown variable if the other two are known.
Image Distance
Image distance, indicated as \(d_i\), is an important parameter that helps you locate an image formed by a lens. It refers to the distance between the lens and the image itself. Image distance can serve as a guide to understanding whether an image is real or virtual.
  • If \(d_i\) is positive, it signifies that the image forms on the side opposite the object, suggesting a real image.
  • If \(d_i\) is negative, the image forms on the same side as the object, indicating a virtual image.
In our exercise, calculating the image distance for different object distances was key:Using the lens formula: \[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]For the object initially at 1.30 m, \(d_{i1} \) was calculated as 0.0939 m or 93.9 mm.
When the object was moved to 6.50 m, \(d_{i2}\) became 0.0909 m or 90.9 mm.
This change in image distance showed us the need to adjust the sensor's position by 3.0 mm.
Focal Length
Focal length is a fundamental property of any lens, defining its converging or diverging capability and determining how it focuses light.
It is denoted by \(f\) in the lens formula and has a significant impact on the formation and characteristics of the image produced.
  • For a converging lens (like the one in this exercise), the focal length is positive, which helps converge light rays to a point on the opposite side of the lens.
  • A shorter focal length implies a lens with greater power, meaning it bends light more sharply.
  • In our case, the focal length of 90 mm signifies a lens that can capture shorter-ranged imagery effectively.
The focal length is essential for adjusting focus when the object distance changes. As seen in our example, the fixed focal length of 90 mm dictated how much the lens-sensor distance needed to be altered when refocusing.

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Most popular questions from this chapter

In one form of cataract surgery the person's natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the person's eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of 24 cm?

A person swimming 0.80 m below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of 5.20 m above the swimmer. What is the actual height of the diving board above the surface of the water?

What is the size of the smallest vertical plane mirror in which a woman of height \(h\) can see her full-length image?

Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens.

A frog can see an insect clearly at a distance of 10 cm. At that point the effective distance from the lens to the retina is 8 mm. If the insect moves 5 cm farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? (a) 0.02 cm, toward the retina; (b) 0.02 cm, away from the retina; (c) 0.06 cm, toward the retina; (d) 0.06 cm, away from the retina.

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