Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 60

Use the thin-lens equation to show that the magnification for a thin lens is determined by its focal length and the object distance and is given by \(m=f /\left(f-d_{\mathrm{o}}\right)\)

Short Answer

Expert verified
To show that the magnification for a thin lens is determined by its focal length and the object distance and is given by \(m = \frac{f}{f - d_o}\), we used the Thin-Lens Equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Rearranging it to isolate the image distance, we found that \(d_i = \frac{f d_o}{f - d_o}\). Then, by expressing magnification as \(m =\frac{d_i}{d_o}\) and substituting the expression for \(d_i\) in terms of \(f\) and \(d_o\), we derived the final magnification equation: \(m = \frac{f}{f - d_o}\).

Step by step solution

01

Apply the Thin-Lens Equation

Recall the Thin-Lens Equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where: - \(f\) is the focal length of the lens - \(d_o\) is the object distance (the distance between the object and the lens) - \(d_i\) is the image distance (the distance between the lens and the image) Now, we need to express the image distance \(d_i\) in terms of the focal length \(f\) and object distance \(d_o\).
02

Solve for the image distance

Rearrange the Thin-Lens Equation to isolate the image distance \(d_i\): \[ d_i = \frac{f d_o}{f - d_o} \]
03

Express the magnification in terms of object and image distances

Recall that the magnification \(m\) is the ratio of the image distance to the object distance: \[ m = \frac{d_i}{d_o} \] Substitute the expression for \(d_i\) from Step 2 into the magnification equation: \[ m = \frac{\frac{f d_o}{f - d_o}}{d_o} \]
04

Simplify the magnification equation

By dividing through the fractions, we can simplify the magnification equation: \[ m = \frac{f d_o}{(f - d_o) d_o} \] \[ m = \frac{f}{f - d_o} \] Now we've derived a magnification equation for a thin lens in terms of its focal length and the object distance. As required, the formula is: \[ m = \frac{f}{f - d_o} \]

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

Some telephoto cameras use a mirror rather than a lens. What radius of curvature mirror is needed to replace a \(800 \mathrm{mm}-\) focal length telephoto lens?

A small telescope has a concave mirror with a 2.00 -m radius of curvature for its objective. Its eyepiece is a 4.00 cm-focal length lens. (a) What is the telescope's angular magnification? (b) What angle is subtended by a 25,000 km- diameter sunspot? (c) What is the angle of its telescopic image?

Ray tracing for a flat mirror shows that the image is located a distance behind the mirror equal to the distance of the object from the mirror. This is stated as \(d_{i}=-d_{\mathrm{o}}\) since this is a negative image distance (it is a virtual image). What is the focal length of a flat mirror?

For a normal, relaxed eye, a magnifying glass produces an angular magnification of \(4.0 .\) What is the largest magnification possible with this magnifying glass?

A myopic person sees that her contact lens prescription is - 4.00 D. What is her far point?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Thermodynamics

Read Explanation

Medical Physics

Read Explanation

Space Physics

Read Explanation

Particle Model of Matter

Read Explanation

Rotational Dynamics

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