Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 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 -10.4 cm.

Step by step solution

01

Understanding the Lens Setup

First, identify the positions and roles of both lenses. Initially, a converging lens forms a focused image 30.0 cm to the right of it. A diverging lens is then placed 15.0 cm to the right of the converging lens.
02

Determine New Image Distance for Diverging Lens

After placing the diverging lens, the screen must be moved 19.2 cm further to the right, making the total image distance from the converging lens \( 30.0 \, \text{cm} + 19.2 \, \text{cm} = 49.2 \, \text{cm} \). The distance of the image formed by the diverging lens is from its own position, so it is \\( 49.2 \, \text{cm} - 15.0 \, \text{cm} = 34.2 \, \text{cm} \). Thus, the new image distance for the diverging lens is \(34.2 \, \text{cm}\) to the right.
03

Apply Lens Formula for Diverging Lens

Use the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) for the diverging lens, where \( v = 34.2 \, \text{cm} \) (image distance) and \( u = -15.0 \, \text{cm} \) (since the object for diverging lens is virtual, to the left). Solve for \( f \), the focal length of the diverging lens:\[\frac{1}{f} = \frac{1}{34.2} - \frac{1}{-15.0}.\]Calculate this to find \( f \).
04

Calculation

Calculate the reciprocal values:\[\frac{1}{f} = \frac{1}{34.2} + \frac{1}{15.0} \approx 0.0292 + 0.0667 = 0.0959.\]Thus, solve for \( f \):\[f = \frac{1}{0.0959} \approx 10.4 \, \text{cm}.\] The focal length of the diverging lens is \(-10.4 \, \text{cm}\) since it is a diverging lens and should be negative.

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 Vaia!

Key Concepts

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

Converging Lens
A converging lens, often known as a convex lens, is a type of lens that brings light rays together to focus at a single point. When an object is placed in front of a converging lens, light rays from the object converge to form an image. This image can be projected onto a screen if it is real.
A hallmark of converging lenses is that they can create both real and virtual images. These lenses are commonly used in devices like cameras, glasses, and telescopes.
To analyze how they form images, we often use the lens formula, which assists in predicting the position and size of the image formed.
Diverging Lens
In contrast to converging lenses, a diverging lens, or concave lens, spreads out light rays. This lens causes parallel rays of light to diverge away from each other upon passing through the lens.
Diverging lenses always create virtual, upright, and reduced images. They are often used in optical devices like peepholes and eyeglasses for correcting nearsightedness.
Understanding how a diverging lens affects light can be crucial in solving optics problems. In a combined lens system exercise, like the one given, it is essential to analyze the systemic impact both lenses have on the image accurately.
Lens Formula
The lens formula is a fundamental equation in optics that helps us find the relationship between the object distance (\( u \)), the image distance (\( v \)), and the focal length (\( f \)) of a lens. Mathematically, it is represented as:
  • \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
This formula is indispensable when determining properties of the image formed by lenses.
For converging lenses, the focal length is positive, while for diverging lenses, it is negative. In the context of the exercise, the lens formula was used to determine the focal length of a diverging lens when combined with a converging lens.
By substituting the known values into the lens formula, we can calculate the unknown values, which allows us to understand how the image changes with different lens setups.
Optics Problem Solving
Optics problem solving often involves mathematical and conceptual understanding of how lenses and light interact. Let's breakdown the steps that are usually involved:
  • Understand the problem: Clearly identify the type of lenses involved and their arrangements.
  • Use known values: Measure or use given distances for object and images, and apply to relevant formulas.
  • Apply the lens formula: This assists in finding unknown distances or focal lengths when some variables are known.
  • Solve for unknowns: Perform calculations carefully to solve for unknown values like focal length.
In exercises like these, patience and practice are necessary. By making step-by-step analyses and calculations, one can determine how different lenses work together. It is crucial to recognize how each step contributes to finding the correct solution, ensuring a comprehensive understanding of the problem.

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

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is 45.0 cm from his eyes instead of the usual 25.0 cm. (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

(a) You want to use a lens with a focal length of 35.0 cm to produce a real image of an object, with the height of the image twice the height of the object. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification-what kind of lens do you need, and where should the object be placed?

To determine whether a frog can judge distance by means of the amount its lens must move to focus on an object, researchers covered one eye with an opaque material. An insect was placed in front of the frog, and the distance that the frog snapped its tongue out to catch the insect was measured with high-speed video. The experiment was repeated with a contact lens over the eye to determine whether the frog could correctly judge the distance under these conditions. If such an experiment is performed twice, once with a lens of power -9-D and once with a lens of power -15-D, in which case does the frog have to focus at a shorter distance, and why? (a) With the -9-D lens; because the lenses are diverging, the lens with the longer focal length creates an image that is closer to the frog. (b) With the -15-D lens; because the lenses are diverging, the lens with the shorter focal length creates an image that is closer to the frog. (c) With the -9-D lens; because the lenses are converging, the lens with the longer focal length creates a larger real image. (d) With the -15-D lens; because the lenses are converging, the lens with the shorter focal length creates a larger real image.

A frog can see an insect clearly at a distance of 10 cm. At that point the effective distance from the lens to the retina is 8 mm. If the insect moves 5 cm farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? (a) 0.02 cm, toward the retina; (b) 0.02 cm, away from the retina; (c) 0.06 cm, toward the retina; (d) 0.06 cm, away from the retina.

A spherical, concave shaving mirror has a radius of curvature of 32.0 cm. (a) What is the magnification of a person's face when it is 12.0 cm to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

See all solutions

Recommended explanations on Physics Textbooks

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