/*! 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 48 A converging lens with a focal l... [FREE SOLUTION] | 91Ó°ÊÓ

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A converging lens with a focal length of \(4.0 \mathrm{cm}\) is to the left of a second identical lens. When a feather is placed \(12 \mathrm{cm}\) to the left of the first lens, the final image is the same size and orientation as the feather itself. What is the separation between the lenses?

Short Answer

Expert verified
The separation between the lenses is 14 cm.

Step by step solution

01

Understand the Lens Equation

The lens equation for a converging lens is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. In this case, we have two identical lenses.
02

Apply the Lens Equation to the First Lens

The object distance \( d_o \) for the first lens is given as 12 cm and \( f = 4.0 \) cm. Substitute these into the lens equation:\[ \frac{1}{4} = \frac{1}{12} + \frac{1}{d_{i1}} \]Solve for \( d_{i1} \), the image distance from the first lens.
03

Calculate \( d_{i1} \)

Rearrange the equation to solve for \( d_{i1} \):\[ \frac{1}{d_{i1}} = \frac{1}{4} - \frac{1}{12} \]Calculate:\[ \frac{1}{d_{i1}} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \]Thus, \( d_{i1} = 6 \text{ cm} \).
04

Consider the Image from the First Lens as the Object for the Second Lens

The image formed by the first lens (\( d_{i1} = 6 \text{ cm} \)) acts as the virtual object for the second lens. The object for the second lens is located on the opposite side of the original object, so the virtual object distance is equal to the separation \( s \) minus 6 cm (the image distance from the first lens).
05

Apply the Lens Equation to the Second Lens

The focal length of the second lens is also 4.0 cm. Let \( d_{i2} \) be the final image distance (where the size and orientation are equal to the original object). The magnification must be 1, indicating that \( d_{o2} = d_{i2} \) (the image distance for the second lens) is the same:\[ \frac{1}{4} = \frac{1}{d_{o2}} + \frac{1}{d_{o2}} = \frac{2}{d_{o2}} \]Thus, \( d_{o2} = 8 \text{ cm} \).
06

Calculate the Separation Between the Lenses

Set the virtual object distance for the second lens, \( s - 6 = d_{o2} \) (virtual object distance). Therefore, \( s - 6 = 8 \text{ cm} \).Solve for \( s \):\[ s = 8 + 6 = 14 \text{ cm} \].
07

Conclusion: Find the Lens Separation

The separation between the two lenses which allows the final image to be the same size and orientation as the original feather is 14 cm.

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Key Concepts

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

Lens Equation
The lens equation is a crucial formula used to analyze the formation of images through lenses, particularly converging lenses. It is expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). This equation relates three important variables:
  • \( f \): the focal length of the lens,
  • \( d_o \): the distance from the object to the lens, known as the object distance,
  • \( d_i \): the distance from the image to the lens, called the image distance.
By rearranging this equation, we can solve for any of these values if the other two are known. For instance, in a scenario involving a feather and two converging lenses, this equation helps determine how the feather's image is manipulated as it encounters each lens. Understanding how to use this equation is fundamental to grasping lens behavior.
Focal Length
The focal length, denoted as \( f \), is a measure of how strongly a lens focuses or diverges light. For a converging lens, this is the distance at which parallel rays of light converge to a single point. A shorter focal length implies a more powerful lens capable of bending light rays more steeply. In the context of our problem, both lenses have a focal length of 4.0 cm. This uniformity plays a significant role in manipulating the image as it passes through both lenses. Knowing the focal length is pivotal because it allows us to apply the lens equation effectively, helping us shape light paths and predict image characteristics.
Image Distance
Image distance, denoted as \( d_i \), is the distance from the lens to the formed image. It is one of the key parameters in the lens equation. When an object is placed at a certain distance from a lens, the image distance tells us where the image will appear relative to the lens. For the first lens in our problem, the calculated image distance, after solving the lens equation, is 6 cm. This means that the image created by the first lens is 6 cm away from it, forming on the side opposite to the incoming light. As we progress to the second lens, the position of this image becomes crucial, as it acts as the virtual object for the next stage of lens interaction.
Object Distance
Object distance, denoted as \( d_o \), is the distance from an object to the lens. It determines where the object is placed in relation to the lens before the image formation begins. In our exercise, the initial feather is placed 12 cm from the first lens. For a converging lens, correctly understanding the object distance is significant because it directly affects the image's characteristics such as size and position. As images progress through subsequent lenses, these distances readjust, leading to new object distances for the next lens stage. For instance, the intermediate image created by the first lens dictates the new object setup for the second lens, guiding us to solve for the separation between the lenses.

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Most popular questions from this chapter

With unaided vision, a physician can focus only on objects that lie at distances between \(5.0 \mathrm{m}\) and \(0.50 \mathrm{m}\). (a) Which type of lens (converging or diverging) is needed to correct her nearsightedness? Explain. (b) Which type of lens will correct her farsightedness? Explain. (c) Find the refractive power needed for each part of the bifocal contact lenses that will give the physician normal visual acuity from \(25 \mathrm{cm}\) out to infinity.

Intracorneal Ring An intracorneal ring is a small plastic device implanted in a person's cornea to change its curvature. By changing the shape of the cornea, the intracomeal ring can correct a person's vision. (a) If a person is nearsighted, should the ring increase or decrease the cornea's curvature? (b) Choose the best explanation from among the following: I. The intracorneal ring should increase the curvature of the cornea so that it bends light more. This will allow it to focus on light coming from far away. II. The intracorneal ring should decrease the curvature of the cornea so it's flatter and bends light less. This will allow parallel rays from far away to be focused.

A 2.05-m-tall object is placed 30.0 cm to the left of a converging lens with a focal length \(f_{1}=20.5 \mathrm{cm} .\) A diverging lens, with a focal length \(f_{2}=-42.5 \mathrm{cm},\) is placed \(30.0 \mathrm{cm}\) to the right of the first lens. How tall is the final image of the object?

The relaxed eyes of a patient have a refractive power of 48.5 diopters. (a) Is this patient nearsighted or farsighted? Explain. (b) If this patient is nearsighted, find the far point. If this person is farsighted, find the near point. (For the purposes of this problem, treat the eye as a single-lens system, with the retina \(2.40 \mathrm{cm}\) from the lens.)

A student has two lenses, one of focal length \(f_{1}=5.0 \mathrm{cm}\) and the other with focal length \(f_{2}=13 \mathrm{cm} .\) (a) When used as a simple magnifier, which of these lenses can produce the greater magnification? Explain. (b) Find the maximum magnification produced by cach of these lenses.

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