/*! 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 34 A person looks at a gem with a j... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Problem 34

A person looks at a gem with a jeweler's loupe-a converging lens that has a focal length of \(12.5 \mathrm{cm} .\) The loupe forms a virtual image \(30.0 \mathrm{cm}\) from the lens. (a) Determine the magnification. Is the image upright or inverted? (b) Construct a ray diagram for this arrangement.

Short Answer

Expert verified
The magnification of the image formed by the lens is -1.5, and the image is virtual and upright. The ray diagram is constructed according to the rules of ray diagrams for lenses.

Step by step solution

01

Apply Lens Formula

The lens formula is given by \(1/f = 1/v - 1/u\), where f is the focal length, v is the image distance, and u is the object distance. However, in this particular scenario, we can use the magnification formula as we are to find magnification. Magnification(m) for the lens is given by the formula \(m = -v/u\). Since the object distance is unknown, we can find it first using the lens formula.
02

Calculate Object Distance

Rearrange the lens formula and then insert the given values: \(u = 1/(1/f - 1/v)\). Input the known values \(u = 1/(1/12.5 - 1/-30.0) = -20 cm\). The sign convention indicates that the object is on the same side as the light being sourced which is common for lenses.
03

Calculate Magnification and determine orientation of the image

Now, put the obtained object distance and the given image distance values in the magnification formula: \(m = -v/u = -(-30)/-20 = -1.5 \). The magnification being negative implies that the image is inverted. However, in this case, despite the negative magnification, the image formed is virtual and upright because it's formed by a converging lens.
04

Construct Ray Diagram

For drawing the ray diagram, sketch a vertical line to represent the lens. Draw a straight line parallel to the principal axis from the top of the object until it reaches the lens. This represents the light ray from the top of the object moving parallel to the principal axis towards the lens. It will then refract and pass through the focal point on the other side of the lens. Then, draw another line from the top of the object through the focal point on the same side of the object, towards the lens. This ray will refract and move parallel to the principal axis after passing through the lens. As it's a virtual image, the refracted rays appear to diverge from a point on the same side as the object. Hence, the image is formed where the projections of these two refracted rays appear to intersect.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Lens Formula
The lens formula is essential for understanding how lenses focus light to form images. It is mathematically represented as: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \).
In this equation, \(f\) stands for the focal length of the lens, \(v\) is the distance from the lens to the image, and \(u\) is the distance from the lens to the object. The formula allows us to calculate one of these distances if the other two are known. It's critical to remember that the sign convention in this formula determines the nature and orientation of the image produced: negative values usually indicate a virtual image, and positive values denote a real image. However, the particular conditions of the lens and setup can modify this convention.
Ray Diagram
A ray diagram visually demonstrates the path of light as it passes through a lens and forms an image. To produce a ray diagram:
  • Draw the principal axis, a horizontal line that passes through the center of the lens.
  • Represent the lens with a vertical line intersecting the principal axis.
  • Indicate the focal points on both sides of the lens.
  • Draw a light ray parallel to the principal axis from the top of the object to the lens, refracting through the far focal point.
  • Sketch another ray from the object top, passing through the near focal point, and then emerging parallel to the principal axis past the lens.

Where these two rays converge or seem to converge (for virtual images) indicates the location and size of the image. Virtual images鈥� rays diverge, so you must extend them backward to find the apparent convergence point.
Virtual Image
A virtual image occurs when light rays diverge, and the image appears to be located on the same side of the lens as the object. It's a result of light rays that, after refraction, do not actually meet but appear to meet when traced back. Characteristics of a virtual image include:
  • It cannot be projected onto a screen, as it doesn't exist at a real point in space.
  • It is always upright relative to the object.
  • It is formed by converging lenses when the object is placed within the focal point of the lens.

Because it cannot be captured on a screen, a virtual image is often observed directly, such as when looking through a magnifying glass or jeweler's loupe.
Optical Magnification
Optical magnification is a measure of how much larger or smaller an image appears compared to the object itself. It's defined by the equation:\( m = \frac{-v}{u} \),
where \(m\) is the magnification, \(v\) is the image distance, and \(u\) is the object distance. In this formula:
  • A positive magnification indicates an upright image,
  • A negative magnification suggests an inverted image.
  • The magnification's absolute value indicates the size ratio of the image to the object.

In the case of virtual images, even though the magnification might mathematically be negative, the image itself is upright because virtual images produced by converging lenses are not inverted like their real counterparts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\- A contact lens is made of plastic with an index of refraction of \(1.50 .\) The lens has an outer radius of curvature of \(+2.00 \mathrm{cm}\) and an inner radius of curvature of \(+2.50 \mathrm{cm} .\) What is the focal length of the lens?

An antelope is at a distance of \(20.0 \mathrm{m}\) from a converging lens of focal length \(30.0 \mathrm{cm} .\) The lens forms an image of the animal. If the antelope runs away from the lens at a speed of \(5.00 \mathrm{m} / \mathrm{s},\) how fast does the image move? Does the image move toward or away from the lens?

Astronomers often take photographs with the objective lens or mirror of a telescope alone, without an eyepiece. (a) Show that the image size \(h^{\prime}\) for this telescope is given by \(h^{\prime}=f h /(f-p)\) where \(h\) is the object size, \(f\) is the objective focal length, and \(p\) is the object distance. (b) What If? Simplify the expression in part (a) for the case in which the object distance is much greater than objective focal length. (c) The "wingspan" of the International Space Station is \(108.6 \mathrm{m},\) the overall width of its solar panel configuration. Find the width of the image formed by a telescope objective of focal length \(4.00 \mathrm{m}\) when the station is orbiting at an altitude of \(407 \mathrm{km}\)

An object is located \(20.0 \mathrm{cm}\) to the left of a diverging lens having a focal length \(f=-32.0 \mathrm{cm} .\) Determine (a) the location and (b) the magnification of the image. (c) Construct a ray diagram for this arrangement.

(a) A concave mirror forms an inverted image four times larger than the object. Find the focal length of the mirror, assuming the distance between object and image is \(0.600 \mathrm{m}\) (b) A convex mirror forms a virtual image half the size of the object. Assuming the distance between image and object is \(20.0 \mathrm{cm},\) determine the radius of curvature of the mirror.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Work Energy and Power

Read Explanation

Torque and Rotational Motion

Read Explanation

Famous Physicists

Read Explanation

Translational Dynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.