Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 61

An object of height \(3.0 \mathrm{cm}\) is placed \(5.0 \mathrm{cm}\) in front of a converging lens of focal length \(20 \mathrm{cm}\) and observed from the other side. Where and how large is the image?

Short Answer

Expert verified
The image is formed at a distance of \(4.0\,\text{cm}\) from the lens and has a height of \(2.4\,\text{cm}\).

Step by step solution

01

Write down the given information

Object height, \(h_o = 3.0\,\text{cm}\) Object distance, \(d_o = 5.0\,\text{cm}\) Focal length of the converging lens, \(f = 20\,\text{cm}\)
02

Apply the lens formula

The lens formula relates the object distance (do), image distance (di), and focal length (f). It is given by: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) Substitute the given values and solve for di: \(\frac{1}{20\,\text{cm}} = \frac{1}{5.0\,\text{cm}} + \frac{1}{d_i}\)
03

Solve for image distance

To find the image distance, we need to solve the equation for \(d_i\): \(\frac{1}{d_i} = \frac{1}{20\,\text{cm}} - \frac{1}{5.0\,\text{cm}} =\frac{1}{4.0\,\text{cm}}\) Now, take the reciprocal to find \(d_i\): \(d_i = 4.0\,\text{cm}\) So, the image distance is \(4.0\,\text{cm}\).
04

Calculate the magnification

The magnification (M) of a lens is the ratio of image height (\(h_i\)) to the object height (\(h_o\)), and it's also equal to the ratio of image distance (\(d_i\)) to object distance (\(d_o\)): \(M = \frac{h_i}{h_o} = \frac{d_i}{d_o}\) Now, substitute the values and solve for \(h_i\): \(M = \frac{4.0\,\text{cm}}{5.0\,\text{cm}} = \frac{4}{5}\) \(h_i = M \times h_o = \frac{4}{5} \times 3.0\,\text{cm} = 2.4\,\text{cm}\)
05

State the final answer

The image is formed at a distance of \(4.0\,\text{cm}\) from the lens and it has a height of \(2.4\,\text{cm}\).

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

Suppose a 200 mm-focal length telephoto lens is being used to photograph mountains \(10.0 \mathrm{km}\) away. (a) Where is the image? (b) What is the height of the image of a \(1000 \mathrm{m}\) high cliff on one of the mountains?

What is the angular magnification of a telescope that has a \(100 \mathrm{cm}\) -focal length objective and a \(2.50 \mathrm{cm}\) -focal length eyepiece?

Use the law of reflection to prove that the focal length of a mirror is half its radius of curvature. That is, prove that \(f=R / 2 .\) Note this is true for a spherical mirror only if its diameter is small compared with its radius of curvature.

A doctor examines a mole with a 15.0 -cm focal length magnifying glass held \(13.5 \mathrm{cm}\) from the mole. (a) Where is the image? (b) What is its magnification? (c) How big is the image of a \(5.00 \mathrm{mm}\) diameter mole?

A nearsighted man cannot see objects clearly beyond \(20 \mathrm{cm}\) from his eyes. How close must he stand to a mirror in order to see what he is doing when he shaves?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation

Turning Points in Physics

Read Explanation

Rotational Dynamics

Read Explanation

Medical Physics

Read Explanation

Fundamentals of 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