/*! 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 106 The near point of a naked eye is... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 106

The near point of a naked eye is \(32 \mathrm{~cm}\). When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of \(16 \mathrm{~cm},\) and is held next to the eye. The enlarged image that is seen is located \(64 \mathrm{~cm}\) from the magnifying glass. Determine the angular size of the image.

Short Answer

Expert verified
The angular size of the image is \(0.180\) radians.

Step by step solution

01

Understanding Angular Size

The angular size of an object when viewed directly through the eye is given by the angle it subtends at the eye. In this problem, this angular size is given to be \(0.060\) radians when viewed from the near point, which is \(32\, \mathrm{cm}\).
02

Define Magnifying Glass Magnification

The magnification \(M\) of a magnifying glass is defined as the ratio of the angle subtended by the image, \(\theta'\), to the angle subtended by the object when placed at the near point, \(\theta_{\mathrm{object}}\). Thus, using \(M = \frac{\theta'}{\theta_{\mathrm{object}}}\).
03

Determine Image Magnification

The formula for magnification using a lens is given by \(M = 1 + \frac{D}{f}\), where \(D\) is the near point distance, and \(f\) is the focal length of the lens. Here, \(D = 32\, \mathrm{cm}\) and \(f = 16\, \mathrm{cm}\), so \(M = 1 + \frac{32}{16} = 3\).
04

Calculate Angular Size of the Image

Using the magnification factor from the magnifying glass \(M = 3\), calculate the angular size of the image \(\theta'\) using the relation \(\theta' = M \times \theta_{\mathrm{object}}\). Since \(\theta_{\mathrm{object}} = 0.060 \) radians, it follows that \(\theta' = 3 \times 0.060 = 0.180 \) radians.

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Key Concepts

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

Angular Size
When we talk about angular size, we refer to the apparent size an object takes up in our vision, measured as an angle. Think of it as how much space it takes up on your retina, the light-sensitive part of the eye. If you've ever looked at the moon and compared it to a smaller object at arm's length, you may have noticed that the angular sizes can appear similar even though the physical sizes are hugely different. In our problem, the object at the near point has an angular size of 0.060 radians. This angle is formed between the edges of the object and your eye, creating a visual field unique to its distance and size.
Magnifying Glass
A magnifying glass is a simple lens meant to make objects appear larger. It works by bending light, focusing it closer to your eye, so that the image is much bigger than what you'd see with the naked eye. This is crucial for examining small details that are hard to see otherwise. The specific attribute of a magnifying glass is its focal length, which in this example is 16 cm. The focal length is the distance over which the lens can focus light, directly affecting how much it can magnify an object.
Magnification
Magnification is the process of enlarging the appearance of an object via optical instruments like lenses. When using a magnifying glass, magnification describes how much bigger the object appears compared to its natural size. In formula terms, it is the ratio of the image鈥檚 angle to the angle of the object seen by the eye alone. The magnification power of our magnifying glass is calculated here as 3, which means the object seen through the glass looks three times larger than without it. This depends on both the near point distance (32 cm) and the focal length (16 cm).
Focal Length
The focal length of a lens is a fundamental concept in optics. It refers to the distance from the lens where light rays converge to a single point. A shorter focal length means a lens can provide more powerful magnification, as it bends light more sharply. For our magnifying glass with a 16 cm focal length, this means it focuses light relatively close, thereby enlarging the visible image. Understanding this distance helps in calculating how much the lens will magnify an object based on its optical power.
Image Calculation
Calculating the angular size of an image when viewed through a magnifying glass involves understanding both magnification and angular size. First, determine the magnification using the formula:
  • \( M = 1 + \frac{D}{f} \)
where \( D \) is the near point distance (32 cm) and \( f \) is the focal length (16 cm), yielding a magnification of 3. To find the new angular size, simply multiply this magnification by the angular size when viewed at the near point:
  • \( \theta' = M \times \theta_{\mathrm{object}} \)
  • \( \theta_{\mathrm{object}} = 0.060 \; \text{radians} \)
Thus, the angular size of the image becomes 0.180 radians, showing the effectiveness of the magnifying glass in making the object appear larger.

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Most popular questions from this chapter

A converging lens \((f=25.0 \mathrm{~cm})\) is used to project an image of an object onto a screen. The object and the screen are \(125 \mathrm{~cm}\) apart, and between them the lens can be placed at either of two locations. Find the two object distances.

A farsighted person can read printing as close as \(25.0 \mathrm{~cm}\) when she wears contacts that have a focal length of \(45.4 \mathrm{~cm}\). One day, however, she forgets her contacts and uses a magnifying glass, as in Figure \(26-40 b\). It has a maximum angular magnification of 7.50 for a young person with a normal near point of \(25.0 \mathrm{~cm}\). What is the maximum angular magnification that the magnifying glass can provide for her?

Mars subtends an angle of \(8.0 \times 10^{-5} \mathrm{rad}\) at the unaided eye. An astronomical telescope has an eyepiece with a focal length of \(0.032 \mathrm{~m}\). When Mars is viewed using this telescope, it subtends an angle of \(2.8 \times 10^{-3} \mathrm{rad}\). Find the focal length of the telescope's objective lens.

Interactive LearningWare 26.2 at offers a review of the concepts that play roles in this problem. A diverging lens \((f=-10.0 \mathrm{~cm})\) is located \(20.0\mathrm{~cm}\) to the left of a converging lens \((f=30.0 \mathrm{~cm})\). A \(3.00-\mathrm{cm}\) -tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)?

In Section 26.4 it is mentioned that the reflected and refracted rays are perpendicular to each other when light strikes the surface at the Brewster angle. This is equivalent to saying that the angle of reflection plus the angle of refraction is \(90^{\circ} .\) Using Snell's law and Brewster's law, prove that the angle of reflection plus the angle of refraction is \(90^{\circ} .\)
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