/*! 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 99 When an object is placed at the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 99

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 \(\mathrm{cm}\) to the right of the lens. A diverging lens is now placed 15.0 \(\mathrm{cm}\) to the right of the converging lens, and it is found that the screen must be moved 19.2 \(\mathrm{cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

Short Answer

Expert verified
The focal length of the diverging lens is approximately \(-21.55 \, \text{cm}\).

Step by step solution

01

Identify Given Values

Converging lens: Image distance \( s' = 30.0 \, \text{cm} \). Diverging lens is placed 15.0 cm to the right of the converging lens. The screen is moved an additional 19.2 cm. Therefore, the final image distance for the diverging lens \( s_2' = 30.0 + 19.2 = 49.2 \, \text{cm} \).
02

Calculate Image Distance from Diverging Lens

The light travels 15.0 cm from the first lens to the diverging lens, so the object distance for the diverging lens \( s_2 = 30.0 - 15.0 = 15.0 \, \text{cm} \).
03

Use Lens Formula for Diverging Lens

Use the lens formula \( \frac{1}{f} = \frac{1}{s_2} + \frac{1}{s_2'} \) to find the focal length \( f \) of the diverging lens:\[\frac{1}{f} = \frac{1}{15.0} + \frac{1}{-49.2}\]This needs to be calculated.
04

Perform Calculation

Calculate the right-hand side of the equation:\[\frac{1}{15.0} = 0.0667, \quad \frac{1}{-49.2} = -0.0203\]Now add the fractions:\[\frac{1}{f} = 0.0667 - 0.0203 = 0.0464\]So, \( f = \frac{1}{0.0464} \approx 21.55 \, \text{cm} \).
05

Determine the Focal Length Sign

Since it's a diverging lens, the focal length \( f \) is negative. So, \( f = -21.55 \, \text{cm} \).

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.

Converging lens
A converging lens bends light rays towards each other. It collects rays from an object and focuses them on a single point. This lens is thicker in the middle than at the edges. This thickness allows it to converge light more effectively.

Key properties of a converging lens include:
  • It forms real and inverted images when the object is placed outside the focal point.
  • When the object is between the focal point and the lens, the image appears virtual, upright, and larger.
  • It is used in applications like cameras, projectors, and in correcting farsightedness.
Understanding these properties helps reveal how a converging lens behaves in various optical systems. The ability to focus light onto a specific point is vital for creating clear and sharp images on a screen or sensor.
Diverging lens
A diverging lens spreads light rays outward, as if they originated from a focal point behind the lens. These lenses are thinner in the middle and thicker at the edges. Their design causes parallel rays of light to appear as if they are coming from a single point behind the lens

Characteristics of a diverging lens include:
  • It always forms virtual images that are upright and smaller than the actual object.
  • They are used in glasses to correct nearsightedness, where they help to diverge light before it reaches the eye.
  • As seen in the provided exercise, the focal length of a diverging lens is negative, indicating that the imaginary focal point lies on the same side as the object.
By grasping these concepts, one can effectively predict how diverging lenses modify light paths in practical scenarios.
Focal length
Focal length is the distance from the center of a lens to the focal point. For a converging lens, the focal point is where light rays converge. In diverging lenses, the focal point is where rays appear to diverge from.

Notable aspects of focal length include:
  • Measured in centimeters (or meters) and significantly influences the image formation of any lens.
  • A positive focal length denotes a converging lens, while a negative value signifies a diverging lens.
  • Focal length determines the strength of the lens: a shorter focal length indicates a stronger lens with greater bending power.
Employing the lens formula, \( \frac{1}{f} = \frac{1}{s} + \frac{1}{s'} \), where \( s \) is the object distance and \( s' \) is the image distance, allows for calculating the focal length precisely. Having a clear understanding of how the focal length interacts within an optical system provides detailed insights into optical behaviour and image clarity.

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

Virtual Object. If the light incident from the left onto a convex mirror does not diverge from an object point but instead converges toward a point at a (negative) distance \(s\) to the right of the mirror, this point is called a virtual object. (a) For a convex mirror having a radius of curvature of \(24.0 \mathrm{cm},\) for what range of virtual-object positions is a real image formed? (b) What is the orientation of this real image? (c) Draw a principal- ray diagram showing the formation of such an image.

A photographer takes a photograph of a Boeing 747 airliner (length 70.7 \(\mathrm{m} )\) when it is flying directly overbead an altitude of 9.50 \(\mathrm{km}\) . The lens has a focal length of 5.00 \(\mathrm{m}\) . How long is the image of the airliner on the film?

An object is 16.0 \(\mathrm{cm}\) to the left of a lens. The lens forms an image 36.0 \(\mathrm{cm}\) to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 \(\mathrm{mm}\) tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

A camera lens has a focal length of 180.0 \(\mathrm{mm}\) and an aperture diameter of 16.36 \(\mathrm{mm}\) (a) What is the \(f\) -number of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{30} \mathrm{s}\) at \(f / 11\) , what is the correct exposure at \(f / 2.8 ?\)

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Dynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electrostatics

Read Explanation

Radiation

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.