/*! 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 88 A screen is placed a distance \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 88

A screen is placed a distance \(d\) to the right of an object. A converging lens with focal length \(f\) is placed between the object and the screen. In terms of \(f\), what is the smallest value \(d\) can have for an image to be in focus on the screen?

Short Answer

Expert verified
The smallest distance \(d\) can have is \(4f\).

Step by step solution

01

Understand the Lens Formula

The lens formula relates object distance (\(u\)) to image distance (\(v\)) and focal length (\(f\)):\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]where \(u + v = d\). We need to find the condition for the smallest \(d\) for which an image is in focus.
02

Express the Relationship Between Distances

Since the object and image distances add up to \(d\) (i.e., \(u + v = d\)),substitute for \(u\):\[u = d - v\]. Substitute \(u = d - v\) back into the lens formula:\[\frac{1}{f} = \frac{1}{d - v} + \frac{1}{v}\]and simplify to get the condition for \(d\).
03

Solve for the Image Distance

From the previous step:\[\frac{1}{f} = \frac{1}{d-v} + \frac{1}{v} = \frac{v + (d - v)}{v(d-v)}\]Simplifying,\[\frac{1}{f} = \frac{d}{v(d-v)}\]Rearranging, we find:\[fd = v(d-v)\].
04

Determine Condition for Minimum d

Substitute \(v = m\) (the image distance) and rearrange:\[fd = m(d - m)\]\[f = \frac{md - m^2}{d}\]The goal is to minimize \(d\). Recognize the condition for the minimum involves the derivative and \(d \geq 4f\). Hence, \(m = u = v\) when \(d\) is minimized for an image in focus.
05

Calculate the Minimum Distance

Finding minimum \(d\), substitute \(v = d/2\) (since it gives the smallest product for a given sum):\[f = \left( \frac{d/2 \cdot d/2}{d} \right)\]Therefore, \[d = 4f\].

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Converging Lens
A converging lens is a transparent optical device that bends light rays towards each other as they pass through it. This type of lens is thicker at the center than at the edges. Converging lenses are crucial in various optical applications like cameras and eyeglasses because they can focus light to form clear images.

When parallel rays of light hit a converging lens, they get refracted in such a way that they meet at a point called the focus. This behavior is highly useful in focusing images on specific planes, such as a screen or sensor, which we often encounter in practical scenarios.
Focal Length
The focal length of a lens is the distance from the lens to the focus, where the refracted light rays meet. The value of the focal length determines how strongly the lens converges or diverges light.

For a converging lens, the focal length is positive and is an essential parameter in the lens formula: \[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]where \(f\) is the focal length. The focal length directly impacts where the image appears in relation to the lens, whether it's close or far on the other side of the lens.
Object Distance
The object distance is the distance from the object to the lens. It's a critical component in the lens formula as it helps determine the location of the formed image.

Given the equation:\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]we can see that object distance, represented by \(u\), plays a fundamental role in whether the image will form on the screen or not. In practical scenarios, altering the object distance while keeping the focal length constant can help bring the image into proper focus.
Image Distance
Image distance refers to the distance between the lens and the image it forms. Like object distance, it is a crucial part of the lens formula.
  • When the image distance \(v\) is positive, it indicates a real image formed on the opposite side of the object.
  • Real images can be projected onto a screen, which is common when using converging lenses.
The calculation of the image distance becomes particularly important in determining how far the screen should be placed, allowing for clear, focused images. In the solution above, rearranging the lens formula provides insights into achieving the minimum required distance \(d\), which is key to ensuring the image is in focus on the screen.

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

In one form of cataract surgery the person's natural lens, which has become cloudy, is replaced by an artificial lens. The refracting properties of the replacement lens can be chosen so that the person's eye focuses on distant objects. But there is no accommodation, and glasses or contact lenses are needed for close vision. What is the power, in diopters, of the corrective contact lenses that will enable a person who has had such surgery to focus on the page of a book at a distance of 24 cm?

A concave mirror has a radius of curvature of 34.0 cm. (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33), what is its focal length?

(a) For a lens with focal length f, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)

You are in your car driving on a highway at 25 m\(/\)s when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 m\(/\)s when the truck is 2.0 m from the mirror, what is the speed of the truck relative to the highway?

A camera with a 90-mm-focal-length lens is focused on an object 1.30 m from the lens. To refocus on an object 6.50 m from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Energy Physics

Read Explanation

Particle Model of Matter

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Solid State Physics

Read Explanation

Classical Mechanics

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.