/*! 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 45 A camera lens has a focal length... [FREE SOLUTION] | 91Ó°ÊÓ

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A camera lens has a focal length of 200 mm. How far from the lens should the subject for the photo be if the lens is 20.4 cm from the sensor?

Short Answer

Expert verified
The subject should be 10.2 meters from the lens.

Step by step solution

01

Convert Units

First, make sure all units are consistent. The focal length is given as 200 mm, and the lens is 20.4 cm from the sensor. Convert 20.4 cm to mm by multiplying by 10: \[ 20.4 \text{ cm} = 204 \text{ mm} \]
02

Apply the Lens Formula

Use the lens formula which relates the object distance \( u \), the image distance \( v \), and the focal length \( f \): \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]Here, \( f = 200 \text{ mm} \) and \( v = 204 \text{ mm} \). We need to find \( u \).
03

Derive the Object Distance Formula

Re-arrange the lens formula to solve for \( \frac{1}{u} \): \[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]
04

Substitute Values

Substitute the known values into the re-arranged formula: \[ \frac{1}{u} = \frac{1}{200} - \frac{1}{204} \]
05

Calculate \( \frac{1}{u} \)

Perform the calculation of \( \frac{1}{u} \):\[ \frac{1}{u} = \frac{1}{200} - \frac{1}{204} = \frac{204 - 200}{200 \times 204} \]Simplify: \[ \frac{1}{u} = \frac{4}{40800} \]
06

Find Object Distance \( u \)

Invert \( \frac{1}{u} \) to find \( u \): \[ u = \frac{40800}{4} = 10200 \text{ mm} \]
07

Convert to Desired Units

The solution requires the object distance in meters. Convert \( 10200 \text{ mm} \) to meters by dividing by 1000:\[ u = \frac{10200}{1000} = 10.2 \text{ m} \]

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Key Concepts

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

Focal Length
The term "focal length" refers to the distance between the lens and its focus, where parallel light rays converge after passing through the lens. In the context of a camera lens, the focal length is crucial because it determines how zoomed in or out an image appears. A long focal length stretches out the vision, offering a zoomed-in view of a distant object.

In simple terms, the focal length is how "strong" or "powerful" a lens is. It's measured in millimeters (mm). For our particular problem, the focal length is 200 mm. This number indicates how far the sensor captures the image, in terms of its depth of field, and affects the perspective of the photo.

When applying the lens formula in the problem, the focal length becomes part of the equation:
  • Formula: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
  • Given: \( f = 200 \text{ mm} \)
This helps calculate either the object's distance from the lens or the image's distance formed on the sensor, depending mostly on the setup.
Object Distance
Object distance refers to the gap between the object being photographed (or observed) and the lens. In order to get a clear image, understanding and calculating this distance is crucial. For photographers, this is often adjustable depending on the lens and camera settings, allowing for flexibility in taking sharp images.

In mathematical terms, the object distance is represented by \( u \) in the lens formula. The equation can be adjusted to solve for \( u \), which helps determine precisely how far the object should be from the lens to capture a focused photograph. The formula re-arranged to solve for object distance is:
  • \( \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \)
By plugging the numbers into the formula:
  • \( \frac{1}{u} = \frac{1}{200} - \frac{1}{204} \)
  • Resulting in \( u = 10200 \text{ mm} \) (or \( 10.2 \text{ m} \) when converted)
Therefore, by understanding this distance, you can precisely adjust lens settings for the perfect shot.
Image Distance
Image distance measures how far the focused image is from the lens itself, most often ending up on the sensor in a camera setup. This is crucial in determining the configuration of the lens relative to the sensor for achieving a sharp image on the film or digital sensor.

In many cases, like in our problem, the image distance \( v \) is initially provided or can be measured directly if the camera or device has a fixed position. Here, the image distance is given as 204 mm. This value tells us how far the sensor is from the lens, which is crucial for setting up the equation:
  • \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
  • Given: \( v = 204 \text{ mm} \)
By knowing the image distance, along with the focal length, you can employ the lens formula to solve for the required object distance, enabling you to take sharp, focused photographs with precision in how the image appears on the capturing sensor.

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Most popular questions from this chapter

A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice \((n = 1.3092)\). What is its apparent depth when viewed at normal incidence?

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?

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 3.50 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

The focal length of a simple magnifier is 8.00 cm. Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 cm in front of her eye? (b) If the object is 1.00 mm high, what is the height of its image formed by the magnifier?

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