/*! 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 38 A lensmaker wants to make a magn... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 38

A lensmaker wants to make a magnifying glass from glass that has an index of refraction \(n=1.55\) and a focal length of \(20.0 \mathrm{~cm}\). If the two surfaces of the lens are to have equal radii, what should that radius be?

Short Answer

Expert verified
The radius of curvature for the lens should be \(40.0 \mathrm{~cm}\)

Step by step solution

01

Simplify The Lensmaker's Equation

Since the two surfaces of the lens have equal radii (r1 = r2 = r), the equation simplifies to \[1/f = 2(n - 1)/r\] because the difference between r and r is zero.
02

Substitute Known Values

Now we can substitute the given values into the equation: \[1/f = 2(n - 1)/r\]. So we get \[1/20.0 = 2(1.55 - 1)/r \].
03

Solve For The Radius r

To solve for the radius r, we can rearrange the equation to isolate r: \[r = 2(1.55 - 1) * 20.0\]

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.

Lensmaker's Equation
The Lensmaker's Equation is a fundamental relation used in optics, particularly in designing lenses. It relates the focal length of a lens to the refractive index of its material and the radii of curvature of its two surfaces. When creating lenses with specific properties, like focal length, this equation helps determine the necessary shape of the lens. For a lens with two surfaces having equal radii, the equation simplifies considerably.
  • The general form of the equation is: \[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
  • In the exercise, since both surfaces have equal radii \((R_1 = R_2 = r)\), the equation simplifies to: \[ \frac{1}{f} = \frac{2(n - 1)}{r} \]
By understanding how changes in the radius \(r\), and index of refraction \(n\) affect the focal length \(f\), we can fine-tune lens construction for various applications.
Index of Refraction
The index of refraction, denoted as \(n\), is a measure that describes how much a wave, such as light, is slowed down as it passes through a medium. It's a dimensionless number that provides insight into the optical properties of materials.
  • In vacuum, the index of refraction is \(1.0\) because light travels without any hindrance.
  • In any other medium, the speed of light is reduced, and thus, the index of refraction increases, leading to values typically greater than 1.
  • For the exercise, the glass used had an index of refraction, \(n = 1.55\). This value indicates the extent to which the glass bends or refracts light passing through it.
Understanding this concept is crucial as it directly impacts how lenses focus light, affecting things like image clarity and magnification.
Focal Length
The focal length of a lens is a key parameter that defines its power in focusing light. It is the distance between the lens and its focal point, where light rays converge after passing through the lens.
  • Shorter focal lengths imply that the lens is stronger at focusing light, which is often desirable for magnifying applications.
  • In contrast, longer focal lengths are common in lenses that need to cover wider scenes, such as those used in photography.
  • For a magnifying glass, a specific focal length of \(20.0\, \, \mathrm{cm}\) was required in the exercise, influencing the lens dimensions needed to achieve this focusing capability.
Recognizing the role of focal length aids in designing lenses to achieve precise optical results.
Radius of Curvature
The radius of curvature of a lens surface is the radius of the sphere from which that surface segment is derived. It plays a crucial role in the lens's ability to converge or diverge light.
  • When the radius of curvature decreases, the curvature of the lens increases, causing more significant refraction of light rays.
  • Lenses with a large radius have surfaces close to being flat, which mean less bending of light.
  • In the problem's scenario, both curved surfaces of the lens had equal radii, simplifying the calculations. The calculated radius needed was \(r = 2(1.55 - 1) \times 20.0\), emphasizing its connection to achieving the desired focal length.
Understanding how the radius of curvature affects light curvature is essential for designing lenses with specific optical properties.

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

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance \(s\) to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance \(s^{\prime}\) and then use Eq. (34.16) to calculate the focal length \(f\) of the lens. But this procedure won't work with a diverging lens - by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object \(20.0 \mathrm{~cm}\) to the left of the lens, the image is \(29.7 \mathrm{~cm}\) to the right of the lens. You then place a diverging lens \(20.0 \mathrm{~cm}\) to the right of the converging lens and measure the final image to be \(42.8 \mathrm{~cm}\) to the right of the converging lens. Suspecting some inaccuracy in measurement, you repeat the lenscombination measurement with the same object distance for the converging lens but with the diverging lens \(25.0 \mathrm{~cm}\) to the right of the converging lens. You measure the final image to be \(31.6 \mathrm{~cm}\) to the right of the converging lens. (a) Use both lens-combination measurements to calculate the focal length of the diverging lens. Take as your best experimental value for the focal length the average of the two values. (b) Which position of the diverging lens, \(20.0 \mathrm{~cm}\) to the right or \(25.0 \mathrm{~cm}\) to the right of the converging lens, gives the tallest image?

A compound microscope has an objective lens with focal length \(14.0 \mathrm{~mm}\) and an eyepiece with focal length \(20.0 \mathrm{~mm}\). The final image is at infinity. The object to be viewed is placed \(2.0 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) What is the distance between the two lenses? (b) Without making the approximation \(s_{1} \approx f_{1},\) use \(M=m_{1} M_{2}\) with \(m_{1}=-s_{1}^{\prime} / s_{1}\) to find the overall angular magnification of the microscope. (c) What is the percentage difference between your result and the result obtained if the approximation \(s_{1} \approx f_{1}\) is used to find \(M ?\)

A concave mirror has a radius of curvature of \(34.0 \mathrm{~cm}\). (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33 ), what is its focal length?

A lens forms a real image that is \(214 \mathrm{~cm}\) away from the object and \(1 \frac{2}{3}\) times its height. What kind of lens is this, and what is its focal length?

A converging lens with a focal length of \(70.0 \mathrm{~cm}\) forms an image of a \(3.20-\mathrm{cm}\) -tall real object that is to the left of the lens. The image is \(4.50 \mathrm{~cm}\) tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Classical Mechanics

Read Explanation

Engineering Physics

Read Explanation

Scientific Method Physics

Read Explanation

Solid State Physics

Read Explanation

Wave Optics

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.