/*! 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 6 An object is located \(10.0 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 6

An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.

Short Answer

Expert verified
**Answer:** 1. Ray tracing for each lens individually 2. Lens equations for each lens

Step by step solution

01

Understand the given exercise and object location

Consider an object located 10.0 cm in front of the first converging lens (Lens 1) with a focal length of 12.0 cm and a second converging lens (Lens 2) 30.0 cm away from Lens 1 with a focal length of 10.0 cm.
02

Ray tracing for Lens 1

For Lens 1: 1. Draw a ray parallel to the principal axis which refracts through the focal point of Lens 1. 2. Draw a ray passing through the center of Lens 1, which remains undeviated. 3. Draw a ray passing through the focal point of Lens 1, which refracts parallel to the principal axis after passing Lens 1. These three rays will converge at a point defining an intermediate image formed by Lens 1.
03

Find the image distance for Lens 1

Use the image formed in step 2 as an object for Lens 2 and repeat the ray tracing process for Lens 2: 1. Draw a ray parallel to the principal axis which refracts through the focal point of Lens 2. 2. Draw a ray passing through the center of Lens 2, which remains undeviated. 3. Draw a ray passing through the focal point of Lens 2, which refracts parallel to the principal axis after passing Lens 2. These rays will converge at a common point where the final image is formed.
04

Lens equations for Lens 1

The lens equation for Lens 1 is given by \(\frac{1}{f_1}=\frac{1}{d_o}+\frac{1}{d_{i1}}\), where: \(f_1 = 12.0\,\mathrm{cm}\) is the focal length of Lens 1, \(d_o = 10.0\,\mathrm{cm}\) is the object distance from Lens 1, \(d_{i1}\) is the distance of the intermediate image formed by Lens 1. Solve for \(d_{i1}\) to get the image distance formed by Lens 1.
05

Lens equations for Lens 2

The lens equation for Lens 2 is given by \(\frac{1}{f_2}=\frac{1}{d_{o2}}+\frac{1}{d_{f}}\), where: \(f_2 = 10.0\,\mathrm{cm}\) is the focal length of Lens 2, \(d_{o2}\) is the object distance from Lens 2, which is equal to the image distance \(d_{i1}\) from Lens 1 plus the distance between the two lenses (30.0 cm), \(d_{f}\) is the final image distance formed by Lens 2. Solve for \(d_{f}\) to get the final image distance formed by the entire lens system.
06

Compare the results from ray tracing and lens equations

Compare image distances obtained through ray tracing and lens equations. If they are in agreement, the final image location obtained is accurate.

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!

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

Two converging lenses, separated by a distance of \(50.0 \mathrm{cm},\) are used in combination. The first lens, located to the left, has a focal length of \(15.0 \mathrm{cm} .\) The second lens, located to the right, has a focal length of \(12.0 \mathrm{cm} .\) An object, \(3.00 \mathrm{cm}\) high, is placed at a distance of \(20.0 \mathrm{cm}\) in front of the first lens. (a) Find the intermediate and final image distances relative to the corresponding lenses. (b) What is the total magnification? (c) What is the height of the final image?
A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.
You have a set of converging lenses with focal lengths $1.00 \mathrm{cm}, 10.0 \mathrm{cm}, 50.0 \mathrm{cm},\( and \)80.0 \mathrm{cm} .$ (a) Which two lenses would you select to make a telescope with the largest magnifying power? What is the angular magnification of the telescope when viewing a distant object? (b) Which lens is used as objective and which as eyepiece? (c) What should be the distance between the objective and eyepiece?
(a) What is the angular size of the Moon as viewed from Earth's surface? See the inside back cover for necessary information. (b) The objective and eyepiece of a refracting telescope have focal lengths \(80 \mathrm{cm}\) and \(2.0 \mathrm{cm}\) respectively. What is the angular size of the Moon as viewed through this telescope?
Suppose the distance from the lens system of the eye (cornea + lens) to the retina is \(18 \mathrm{mm}\). (a) What must the power of the lens be when looking at distant objects? (b) What must the power of the lens be when looking at an object \(20.0 \mathrm{cm}\) from the eye? (c) Suppose that the eye is farsighted; the person cannot see clearly objects that are closer than $1.0 \mathrm{m}$. Find the power of the contact lens you would prescribe so that objects as close as \(20.0 \mathrm{cm}\) can be seen clearly.
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Kinematics Physics

Read Explanation

Rotational Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Nuclear Physics

Read Explanation

Force

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.