Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 55

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?

Short Answer

Expert verified
Answer: The intermediate and final image distances relative to the corresponding lenses are 60 cm and 60 cm respectively. The total magnification is 18, and the height of the final image is 54 cm.

Step by step solution

01

Find the intermediate image distance

Using the thin lens formula, we can find the image distance for the first lens: \(1/f = 1/d_o + 1/d_i\) Where \(d_o\) is the object distance, \(d_i\) is the image distance, and \(f\) is the focal length. Let's plug in the known values for the first lens and solve for the intermediate image distance, \(d_{i1}\): \(1/15 = 1/20 + 1/d_{i1}\) Now, solve for \(d_{i1}\): \(d_{i1} = 60\,\text{cm}\)
02

Image produced by first lens becomes object for second lens

The image produced by the first lens becomes the object for the second lens. We can find the object distance (\(d_{o2}\)) for the second lens by subtracting the distance between the lenses from the image distance \(d_{i1}\). \(d_{o2} = d_{i1} - 50\) \(d_{o2} = 60 - 50\) \(d_{o2} = 10\,\mathrm{cm}\)
03

Find the final image distance for the second lens

Now, let's use the lens formula to find the final image distance, \(d_{i2}\), for the second lens: \(1/f = 1/d_o + 1/d_i\) Plug in the known values for the second lens: \(1/12 = 1/10 + 1/d_{i2}\) Solve for \(d_{i2}\): \(d_{i2} = 60\,\mathrm{cm}\)
04

Find the magnification for each lens

The magnification of each lens can be found using the formula: \(m = -d_i / d_o\) Calculate the magnification for the first lens, \(m_1\): \(m_1 = -d_{i1} / 20\) \(m_1 = -60 / 20\) \(m_1 = -3\) Calculate the magnification for the second lens, \(m_2\): \(m_2 = -d_{i2} / 10\) \(m_2 = -60 / 10\) \(m_2 = -6\)
05

Find the total magnification

To find the total magnification, multiply the magnification of each lens: \(M = m_1 \times m_2\) \(M = (-3) \times (-6)\) \(M = 18\)
06

Find the final image height

To find the height of the final image, multiply the total magnification with the object's height: \(h_{i} = M \times h_o\) \(h_{i} = 18 \times 3\) \(h_{i} = 54\,\mathrm{cm}\) #Summary# The intermediate and final image distances relative to the corresponding lenses are \(d_{i1} = 60\,\mathrm{cm}\) and \(d_{i2} = 60\,\mathrm{cm}\), respectively. The total magnification is \(18\), and the height of the final image is \(54\,\mathrm{cm}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Kim says that she was less than 10 ft away from the president when she took a picture of him with her \(50-\mathrm{mm}\) focal length camera lens. The picture shows the upper half of the president's body (or \(3.0 \mathrm{ft}\) of his total height). On the negative of the film, this part of his body is $18 \mathrm{mm}$ high. How close was Kim to the president when she took the picture?
The area occupied by one frame on 35 -mm film is \(24 \mathrm{mm}\) by $36 \mathrm{mm}-\( see the figure with Problem \)16 .$ The focal length of the camera lens is \(50.0 \mathrm{mm}\). A picture is taken of a person $182 \mathrm{cm}$ tall. What is the minimum distance from the camera for the person to stand so that the image fits on the film? Give two answers; one for each orientation of the camera.
A dissecting microscope is designed to have a large distance between the object and the objective lens. Suppose the focal length of the objective of a dissecting microscope is \(5.0 \mathrm{cm},\) the focal length of the eyepiece is \(4.0 \mathrm{cm},\) and the distance between the lenses is $32.0 \mathrm{cm} .$ (a) What is the distance between the object and the objective lens? (b) What is the angular magnification?
A camera with a 50.0 -mm lens can focus on objects located from $1.5 \mathrm{m}$ to an infinite distance away by adjusting the distance between the lens and the film. When the focus is changed from that for a distant mountain range to that for a flower bed at \(1.5 \mathrm{m},\) how far does the lens move with respect to the film?
A camera lens has a fixed focal length of magnitude \(50.0 \mathrm{mm} .\) The camera is focused on a 1.0 -m-tall child who is standing \(3.0 \mathrm{m}\) from the lens. (a) Should the image formed be real or virtual? Why? (b) Is the lens converging or diverging? Why? (c) What is the distance from the lens to the film? (d) How tall is the image on the film? (e) To focus the camera, the lens is moved away from or closer to the film. What is the total distance the lens must be able to move if the camera can take clear pictures of objects at distances anywhere from \(1.00 \mathrm{m}\) to infinity?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Force

Read Explanation

Electromagnetism

Read Explanation

Energy Physics

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.

Sign-up for free