/*! 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 112 A camper is trying to start a fi... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 112

A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is \(1.39 \times 10^{\circ} \mathrm{m},\) and its mean distance from the earth is \(1.50 \times 10^{11} \mathrm{m} .\) The camper is using a converging lens whose focal length is \(10.0 \mathrm{cm}\). (a) What is the area of the sun's image on the paper? (b) If \(0.530 \mathrm{W}\) of sunlight passes through the lens, what is the intensity of the sunlight at the paper?

Short Answer

Expert verified
(a) The area of the sun's image is approximately \(6.76 \times 10^{-7}\) m虏. (b) The intensity of the sunlight at the paper is about \(7.84 \times 10^5\) W/m虏.

Step by step solution

01

Understand the Exercise

The exercise involves using optics to focus sunlight onto paper. It requires calculation of the area of the sun's image formed by a lens and the intensity of sunlight on the paper.
02

Determine the Angular Size of the Sun

The angular size of an object is given by the formula \( \theta = \frac{d}{D} \), where \( d \) is the diameter of the object, and \( D \) is the distance. For the sun: \( \theta = \frac{1.39 \times 10^9}{1.50 \times 10^{11}} \approx 0.00927 \) radians.
03

Find Image Diameter Using Lens Equation

The image diameter formed by a lens is the product of the angular size of the object and the focal length of the lens. Image diameter \( d_i = \theta \times f = 0.00927 \times 0.10 \approx 0.000927 \) meters.
04

Calculate Area of Sun's Image

The area \( A \) of a circle is calculated using \( A = \pi \times \left(\frac{d_i}{2}\right)^2 \). Substituting the image diameter: \( A \approx \pi \times \left(\frac{0.000927}{2}\right)^2 \approx 6.76 \times 10^{-7} \) square meters.
05

Figure Out Intensity of Sunlight

Intensity \( I \) is defined as power per unit area. Use \( I = \frac{P}{A} \) where \( P = 0.530 \) watts and \( A = 6.76 \times 10^{-7} \). So, \( I = \frac{0.530}{6.76 \times 10^{-7}} \approx 7.84 \times 10^5 \) watts per square meter.

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Key Concepts

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

Angular Size
Angular size is a measure of how large an object appears to be when seen from a particular point of view. This is particularly useful in astronomy to understand how we perceive the size of celestial objects like the sun and the moon from Earth. The formula to calculate angular size is:
  • \( \theta = \frac{d}{D} \)
where:
  • \( \theta \) is the angular size in radians,
  • \( d \) is the actual diameter of the object, and
  • \( D \) is the distance of the object from the observer.
For example, the sun's angular size as seen from Earth is about 0.00927 radians, derived from its diameter of \(1.39 \times 10^9\) meters and the distance \(1.50 \times 10^{11}\) meters. This relatively small radian value illustrates why the sun appears as a small disk in the sky rather than its actual enormous size.
Converging Lens
A converging lens, also known as a convex lens, bends light rays towards each other, causing them to meet at a point known as the focus. Converging lenses are commonly used in devices such as magnifying glasses, cameras, and of course, for focusing sunlight in our typical campfire exercise. Key characteristics of a converging lens include:
  • A thick center and thin edges,
  • Ability to converge parallel rays of light to a single focal point,
  • A focal length which is the distance from the lens center to the focus.
In the exercise, a focal length of 10 cm is used, which helps in forming a small, intense image of the sun. This intense light can increase temperature at the focal point, enabling the paper to catch fire.
Intensity of Sunlight
Intensity refers to the power per unit area, and when discussing sunlight, it's an expression of how much solar power is hitting a certain area. It can be calculated using the equation:
  • \( I = \frac{P}{A} \)
where:
  • \( I \) is the intensity in watts per square meter,
  • \( P \) is the power passing through the lens, and
  • \( A \) is the area over which this power is distributed.
In the campfire exercise, the power is given as 0.530 watts, and the area of the focused sunlight is calculated to be approximately \(6.76 \times 10^{-7}\) square meters. Plugging these into our formula, we find the intensity is around \(7.84 \times 10^5\) watts per square meter, significantly increasing the chances of igniting the paper.
Lens Equation
The lens equation is a fundamental concept in optics, used to determine the relationship between the object distance, the image distance, and the focal length of a lens. It can be expressed as:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
where:
  • \( f \) 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 not directly applied in the campfire problem, but it underlies the process of focusing light to form clear images on a screen or any surface. A lens with a particular focal length, like the 10 cm one in the exercise, uses this principle to create a sharp image of the sun on the paper, allowing us to calculate its size and concentration using the angular size.

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Most popular questions from this chapter

A woman can read the large print in a newspaper only when it is at a distance of \(65 \mathrm{cm}\) or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn \(2.0 \mathrm{cm}\) from the eyes), so she can read the newspaper at a distance of \(25 \mathrm{cm}\) from her eyes?

A compound microscope has a barrel whose length is \(16.0 \mathrm{cm}\) and an cyepiece whose focal length is \(1.4 \mathrm{cm} .\) The viewer has a near point located \(25 \mathrm{cm}\) from his eyes. What focal length must the objective have so that the angular magnification of the microscope will be \(-320 ?\)

An object is located \(9.0 \mathrm{cm}\) in front of a converging lens \((f=6.0 \mathrm{cm}) .\) Using an accurately drawn ray diagram, determine where the image is located.

A camera is supplied with two interchangeable lenses, whose focal lengths are 35.0 and \(150.0 \mathrm{mm} .\) A woman whose height is \(1.60 \mathrm{m}\) stands \(9.00 \mathrm{m}\) in front of the camera. What is the height (including sign) of her image on the image sensor, as produced by (a) the \(35.0-\mathrm{mm}\) lens and (b) the \(150.0-\mathrm{mm}\) lens?

A nearsighted patient's far point is \(0.690 \mathrm{m}\) from her eyes. She is able to see distant objects in focus when wearing glasses with a refractive power of -1.50 diopters. What is the distance between her eyes and the glasses?
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