/*! 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 2 A converging lens and a divergin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 2

A converging lens and a diverging lens, separated by a distance of $30.0 \mathrm{cm},$ are used in combination. The converging lens has a focal length of \(15.0 \mathrm{cm} .\) The diverging lens is of unknown focal length. An object is placed \(20.0 \mathrm{cm}\) in front of the converging lens; the final image is virtual and is formed \(12.0 \mathrm{cm}\) before the diverging lens. What is the focal length of the diverging lens?

Short Answer

Expert verified
Answer: The focal length of the diverging lens is -60.0 cm.

Step by step solution

01

Understand the given information

We have a converging lens with a focal length of 15cm and a diverging lens with an unknown focal length. The object is placed 20 cm in front of the converging lens, and the final image is virtual, located 12 cm before the diverging lens. The lenses are separated by 30 cm.
02

Calculate the first image position using the lens formula for the converging lens

To find the position of the image formed by the converging lens, we can use the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) Where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. For the converging lens: \(f=15.0\,cm\), \(d_o=20.0\,cm\) We can now solve for \(d_i\), \(\frac{1}{15.0} = \frac{1}{20.0} + \frac{1}{d_i}\)
03

Solve for the image distance \(d_i\) of the converging lens

We need to solve the equation for \(d_i\): Rearrange and solve for \(d_i\), \(\frac{1}{d_i} = \frac{1}{15.0} - \frac{1}{20.0}\) Find a common denominator and subtract the fractions: \(\frac{1}{d_i} = \frac{1}{60}\) Thus, \(d_i = 60.0\,cm\)
04

Calculate the object distance for the diverging lens

The distance from the converging lens image to the diverging lens will be the object distance for the diverging lens. Since lenses are 30 cm apart, and we have calculated the image distance \(d_i\) as 60 cm Object distance, \(d_o' = d_i - 30.0 = 60.0 - 30.0 = 30.0\,cm\)
05

Determine the image distance of the diverging lens

Given that the final image is virtual and is formed 12 cm before the diverging lens, it means the image distance for the diverging lens will be negative. Therefore, \(d_i'=-12.0\,cm\)
06

Apply the lens formula for the diverging lens to find the focal length

Again we apply the lens formula with the new object and image distances for the diverging lens to find its focal length, \(f'\): \(\frac{1}{f'} = \frac{1}{d_o'} + \frac{1}{d_i'}\) Plug in the values we found: \(\frac{1}{f'} = \frac{1}{30.0} + \frac{1}{-12.0}\)
07

Solve for the focal length of the diverging lens

Solve for \(f'\): \(\frac{1}{f'} = \frac{-1}{60}\) So, \(f' = -60.0\,cm\) The focal length of the diverging lens is \(-60.0\,cm\).

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.

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

You have two lenses of focal length \(25.0 \mathrm{cm}\) (lens 1 ) and $5.0 \mathrm{cm}$ (lens 2 ). (a) To build an astronomical telescope that gives an angular magnification of \(5.0,\) how should you use the lenses (which for objective and which for eyepiece)? Explain. (b) How far apart should they be?

A converging lens with a focal length of \(15.0 \mathrm{cm}\) and a diverging lens are placed \(25.0 \mathrm{cm}\) apart, with the converging lens on the left. A 2.00 -cm-high object is placed \(22.0 \mathrm{cm}\) to the left of the converging lens. The final image is \(34.0 \mathrm{cm}\) to the left of the converging lens. (a) What is the focal length of the diverging lens? (b) What is the height of the final image? (c) Is the final image upright or inverted?
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.
Use the thin-lens equation to show that the transverse magnification due to the objective of a microscope is \(m_{o}=-U f_{o} .\) [Hints: The object is near the focal point of the objective: do not assume it is at the focal point. Eliminate \(p_{0}\) to find the magnification in terms of \(q_{0}\) and $\left.f_{0} . \text { How is } L \text { related to } q_{0} \text { and } f_{0} ?\right].$
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

Particle Model of Matter

Read Explanation

Measurements

Read Explanation

Oscillations

Read Explanation

Fundamentals of Physics

Read Explanation

Magnetism

Read Explanation

Famous Physicists

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.