Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 33: Problem 90

A diverging lens with \(f=-30.0 \mathrm{~cm}\) is placed \(15.0 \mathrm{~cm}\) behind a converging lens with \(f=20.0 \mathrm{~cm}\). Where will an object at infinity in front of the converging lens be focused?

Short Answer

Expert verified
Answer: The object will be focused at a distance of \(5.0~cm\) behind the diverging lens.

Step by step solution

01

Find the image position of object due to the converging lens

Since the object is at infinity, the image will be at the focal point of the converging lens. So, the image position due to the converging lens (\(d_{i1}\)) is \(20.0~cm\) behind it.
02

Find the object position for the diverging lens

The object for the diverging lens will be the image formed by the converging lens. We know that the distance between the two lenses is \(15.0~cm\), so the object is \(20.0 - 15.0 = 5.0~cm\) away from the diverging lens. We can denote this distance as \(d_{o2}\).
03

Use Lensmaker's equation for the diverging lens

Lensmaker's equation is given as follows: \(\frac{1}{f} = \frac{1}{d_o} - \frac{1}{d_i}\) For the diverging lens, \(f = -30.0~cm\) and \(d_{o2} = 5.0~cm\). We will now use this equation to find the image position for the diverging lens (\(d_{i2}\)). \(\frac{1}{-30.0} = \frac{1}{5.0} - \frac{1}{d_{i2}}\)
04

Solve for \(d_{i2}\)

We have the equation: \(\frac{1}{-30.0} = \frac{1}{5.0} - \frac{1}{d_{i2}}\) To solve for \(d_{i2}\), we will first subtract \(\frac{1}{5.0}\) from both sides: \(\frac{1}{-30.0} - \frac{1}{5.0} = -\frac{1}{d_{i2}}\) Now we need to find a common denominator for the fractions on the left-hand side of the equation. The common denominator will be \(30.0\): \(\frac{-5-1}{30.0} = -\frac{1}{d_{i2}}\) \(\frac{-6}{30.0} = -\frac{1}{d_{i2}}\) Now we have fractions with the same numerator, so we can easily find \(d_{i2}\): \(d_{i2} = 30.0 \cdot \frac{1}{6} = 5.0~cm\)
05

Final Answer

The final image position is \(5.0~cm\) behind the diverging lens.

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

The radius of curvature for the outer part of the cornea is \(8.0 \mathrm{~mm}\), the inner portion is relatively flat. If the index of refraction of the cornea and the aqueous humor is 1.34: a) Find the power of the cornea. b) If the combination of the lens and the cornea has a power of \(50 .\) diopter, find the power of the lens (assume the two are touching).

A telescope is advertised as providing a magnification of magnitude 41 using an eyepiece of focal length \(4.0 \cdot 10^{1} \mathrm{~mm}\). What is the focal length of the objective?

An object is \(6.0 \mathrm{~cm}\) from a thin lens along the axis of the lens. If the lens has a focal length of \(9.0 \mathrm{~cm},\) determine the image distance.

A person with a near-point distance of \(24.0 \mathrm{~cm}\) finds that a magnifying glass gives an angular magnification that is 1.25 times larger when the image of the magnifier is at the near point than when the image is at infinity. What is the focal length of the magnifying glass?

A converging lens will be used as a magnifying glass. In order for this to work, the object must be placed at a distance a) \(d_{\mathrm{o}}>f\). c) \(d_{\mathrm{o}}
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Physical Quantities And Units

Read Explanation

Linear Momentum

Read Explanation

Thermodynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Modern 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