/*! 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 117 A converging lens \((f=12.0 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 117

A converging lens \((f=12.0 \mathrm{cm})\) is held 8.00 \(\mathrm{cm}\) in front of a newsaper that has a print size with a height of 2.00 \(\mathrm{mm}\) . Find \((\mathrm{a})\) the image distance (in \(\mathrm{cm} )\) and \((\mathrm{b})\) the height (in mm) of the magnified print.

Short Answer

Expert verified
(a) Image distance: -24 cm; (b) Height of magnified print: 6 mm.

Step by step solution

01

Understand the Lens Formula

A converging lens focuses light such that it creates an image of an object. We use the lens formula to establish a relationship between the object distance (\(d_o\)), the image distance (\(d_i\)), and the focal length (\(f\)) of the lens. The lens formula is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here, \(f = 12.0 \text{ cm}\) and \(d_o = 8.0 \text{ cm}\).

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.

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.

Lens Formula
The lens formula is crucial in understanding how lenses, especially converging ones, operate. This formula relates the object distance (\(d_o\)), image distance (\(d_i\)), and the focal length (\(f\)) of a lens. The mathematical representation is:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]In this scenario, we're dealing with a lens with a focal length of 12.0 cm. The object (a newspaper, in this case) is placed 8.0 cm in front of the lens. By substituting these values into the formula, we can find the unknown, which is the image distance. The lens formula is pivotal in optics as it helps predict where and how the image will form, a fundamental aspect when working with lenses.
Focal Length
The focal length (\(f\)) is a key property of any lens, determining how strongly it converges (or diverges) light. It is the distance from the center of the lens to its focal point, where parallel light rays converge.For a converging lens, such as the one in the exercise, the focal length is positive. A shorter focal length indicates a stronger converging ability, meaning that the lens bends light more sharply. In this exercise, with \(f = 12.0 \text{ cm}\), the lens can create a focused image relatively close to it. Understanding focal length helps in visualizing whether the image will be real or virtual and its relative size.
Image Distance
Image distance (\(d_i\)) tells us how far the resulting image is from the lens. By rearranging the lens formula, we can solve for \(d_i\):\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\]For our exercise, substituting in the known values of \(f = 12.0 \text{ cm}\) and \(d_o = 8.0 \text{ cm}\) allows us to determine the exact location of the image. Understanding whether this distance is positive or negative helps in deciding if the image is real (projected onto a screen) or virtual (only visible by looking through the lens). This calculation is crucial for the correct positioning of objects relative to the lens.
Magnification
Magnification describes how much larger or smaller the image appears compared to the object. The magnification (\(m\)) is calculated using:\[m = \frac{-d_i}{d_o}\]We also find the height of the image using:\[\text{Image height} = m \times \text{Object height}\]In this exercise, the newspaper's print height is given as 2.00 mm. Using the calculated magnification, we can determine the new height of the print as seen through the lens. Magnification provides insight into whether the image is upright or inverted, and whether it's magnified or reduced. It's a direct outcome of the image and object distances, tying all concepts together in optics.

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

An object has an angular size of 0.0150 rad when placed at the near point (21.0 cm) of an eye. When the eye views this object using a magnifying glass, the largest possible angular size of the image is 0.0380 rad. What is the focal length of the magnifying glass?

The distance between an object and its image formed by a diverging lens is 49.0 cm. The focal length of the lens is 233.0 cm. Find (a) the image distance and (b) the object distance.

A scuba diver, submerged under water, looks up and sees sunlight at an angle of \(28.0^{\circ}\) from the vertical. At what angle, measured from the vertical, does this sunlight strike the surface of the water?

The refractive indices of materials \(A\) and \(B\) have a ratio of \(n_{A} / n_{B}=1.33\) . The speed of light in material \(A\) is \(1.25 \times 10^{8} \mathrm{m} / \mathrm{s}\) . What is the speed of light in material \(B ?\)

A compound microscope has a barrel whose length is 16.0 cm and an eyepiece whose focal length is 1.4 cm. The viewer has a near point located 25 cm from his eyes. What focal length must the objective have so that the angular magnification of the microscope will be 320?
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Classical Mechanics

Read Explanation

Modern Physics

Read Explanation

Energy Physics

Read Explanation

Rotational Dynamics

Read Explanation

Particle Model of Matter

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.