/*! 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 122 An object is placed to the left ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 122

An object is placed to the left of a lens, and a real image is formed to the right of the lens. The image is inverted relative to the object and is onehalf the size of the object. (a) What kind of lens, converging or diverging, is used to produce this image? (b) How is the height \(h_{\mathrm{i}}\) of the image related to the height \(h_{0}\) of the object? Don't forget to take into account the fact that the image is inverted relative to the object. (c) What is the ratio \(d_{i} / d_{0}\) of the image distance to the object distance? For the situation described in the Concept Questions, the distance between the object and the image is \(90.0 \mathrm{~cm}\). (a) How far from the lens is the object? (b) What is the focal length of the lens?

Short Answer

Expert verified
(a) Converging lens. (b) \( \frac{h_i}{h_0} = -0.5 \). (c) \( \frac{d_i}{d_0} = 0.5 \). Object is 60 cm from lens, focal length is 20 cm.

Step by step solution

01

Determine the Lens Type

For a real and inverted image formed on the opposite side of the object by a lens, a converging lens is used. Diverging lenses create virtual images that are upright. In this case, given the conditions of the image, the lens must be converging.
02

Relate Image Height to Object Height

The magnification formula relates the heights of the image and the object. It's given by \( m = \frac{h_i}{h_0} \). Since the image is inverted and half the size: \( \frac{h_i}{h_0} = -0.5 \). The negative sign is because the image is inverted relative to the object.
03

Calculate Image Distance to Object Distance Ratio

Since magnification \( m = \frac{-d_i}{d_0} \), substituting the magnification value from the previous step gives \( \frac{-d_i}{d_0} = -0.5 \). Therefore, \( \frac{d_i}{d_0} = 0.5 \).
04

Calculate Object and Image Distance from Lens

Given that the total distance between the object and the image is 90 cm, we apply the relationship from Step 3. If \( d_i = 0.5 \times d_0 \), we have: \( d_0 + d_i = 90 \, \text{cm} \) or \( d_0 + 0.5d_0 = 90 \). This simplifies to \( 1.5d_0 = 90 \), solving for \( d_0 \) gives \( d_0 = 60 \, \text{cm} \). Then \( d_i = 0.5 \times 60 = 30 \, \text{cm} \).
05

Calculate the Focal Length of the Lens

Using the lens formula \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i} \), substitute \( d_0 = 60 \, \text{cm} \) and \( d_i = 30 \, \text{cm} \): \[ \frac{1}{f} = \frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20} \] Thus, \( f = 20 \, \text{cm} \).

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.

Magnification Formula
The magnification formula is a crucial concept when analyzing images formed by lenses. This formula helps you understand how the size of an image relates to the object's size. If you're given the height of the object, denoted as \( h_0 \), and the height of the image as \( h_i \), then the formula can be expressed as:
  • \( m = \frac{h_i}{h_0} \)
Magnification \( m \) provides us with the size relationship between the object and image. In cases where the image is inverted, as it is here, the magnification will be negative. This indicates not only a size change but also a flip. For example, if the magnification is \(-0.5\), the image is half the size of the object and inverted. Understanding this formula helps you determine both the size and orientation of the image formed.
Real and Inverted Image
A real and inverted image is precisely what it sounds like鈥攁n image not only has size, but it also flips upside down relative to the object. This is a common characteristic of images formed by converging lenses. These lenses, also known as convex lenses, are designed to converge light rays to a point. Real images result because the rays of light actually meet at a point after passing through the lens. This is different from virtual images, which appear to diverge from a point behind the lens. For real images, because the light physically converges, we can project these images onto a screen. It's important to note that such images tend to be inverted compared to the object due to the way light rays intersect after passing through the lens.
Focal Length Calculation
Calculating the focal length of a lens is essential for understanding its optical properties. The focal length, denoted \( f \), is the distance from the lens where parallel rays of light converge to a point.We use the lens formula to find \( f \) when object distance \( d_0 \) and image distance \( d_i \) are known:
  • \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i} \)
Let's break it down with a practical example. Suppose an object is placed 60 cm away from the lens, and the image forms 30 cm on the opposite side. Plug these values into the formula:\[\frac{1}{f} = \frac{1}{60} + \frac{1}{30} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20}\]From this calculation, the focal length \( f \) is found to be 20 cm. This distance indicates the point where all parallel light rays focus, showing the lens's strength in converging light.

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

The near point of a naked eye is \(25 \mathrm{~cm}\). When placed at the near point and viewed by the naked eye, a tiny object would have an angular size of \(5.2 \times 10^{-5}\) rad. When viewed through a compound microscope, however, it has an angular size of \(-8.8 \times 10^{-3}\) rad. (The minus sign indicates that the image produced by the microscope is inverted.) The objective of the microscope has a focal length of \(2.6 \mathrm{~cm},\) and the distance between the objective and the eyepiece is \(16 \mathrm{~cm} .\) Find the focal length of the eyepiece.

Two systems are formed from a converging lens and a diverging lens, as shown in parts \(a\) and \(b\) of the drawing. (The point labeled \(" F_{\text {converging }}^{\prime \prime}\) is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 . Without doing any calculations, determine for each system whether the final image lies to the left or to the right of lens \(2 .\) Provide a reason for each answer. The focal lengths of the converging and diverging lenses are \(15.00\) and \(-20.0\) \(\mathrm{cm}\), respectively. The distance between the lenses is \(50.0 \mathrm{~cm}\), and an object is placed \(10.00 \mathrm{~cm}\) to the left of lens \(1 .\) Determine the final image distance for each system, measured with respect to lens 2. Check to be sure your answers are consistent with your answers to the Concept Question.

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text {diamond }}=2.444, n_{\text {crown glass }}=1.531\right),\) and red \(\left(n_{\text {diamond }}=2.410, n_{\text {crown glass }}=1.520\right)\) Determine the angle between the refracted blue and red rays in the crown glass.

Red light \((n=1.520)\) and violet light \((n=1.538)\) traveling in air are incident on a slab of crown glass. Both colors enter the glass at the same angle of refraction. The red light has an angle of incidence of \(30.00^{\circ} .\) What is the angle of incidence of the violet light?

An object is placed \(20.0 \mathrm{~cm}\) to the left of a diverging lens \((f=-8.00 \mathrm{~cm}) .\) A concave mirror \((f=12.0 \mathrm{~cm})\) is placed \(30.0 \mathrm{~cm}\) to the right of the lens. (a) Find the final image distance, mea sured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Quantum Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Oscillations

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Further Mechanics and Thermal Physics

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.