/*! 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 59 The focal length of a simple mag... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 59

The focal length of a simple magnifier is 8.00 \(\mathrm{cm} .\) Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 \(\mathrm{cm}\) in front of her eye? (b) If the object is 1.00 mm high, what is the height of its image formed by the magnifier?

Short Answer

Expert verified
(a) 11.76 cm; (b) 2.13 mm.

Step by step solution

01

Understanding the Lens Formula

The lens formula relates the object distance \( u \), the image distance \( v \), and the focal length \( f \) of a lens. It is given by:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]For a magnifier with the image formed at the near point (\( v = 25.0 \; \text{cm} \)) and focal length \( f = 8.00 \; \text{cm} \), we can use this formula to find the object distance \( u \).
02

Substituting Values into the Lens Formula

Substitute the known values into the lens equation:\[ \frac{1}{8.00} = \frac{1}{25.0} + \frac{1}{u} \]To find \( \frac{1}{u} \), we rearrange the equation:\[ \frac{1}{u} = \frac{1}{8.00} - \frac{1}{25.0} \]
03

Calculating the Object Distance

Calculate \( \frac{1}{u} \) by finding \( \frac{1}{8.00} - \frac{1}{25.0} \):\[ \frac{1}{u} = \frac{25.0 - 8.00}{8.00 \times 25.0} = \frac{17.0}{200.0} \]Thus, \( u = \frac{200.0}{17.0} \approx 11.76 \; \text{cm} \).So, the object should be placed approximately 11.76 cm from the magnifier.
04

Understanding Magnification

Magnification \( M \) is the ratio of the image height \( h_i \) to the object height \( h_o \) and is also given by:\[ M = \frac{v}{u} \]The object is 1.00 mm high, so \( h_o = 1.00 \; \text{mm} \).
05

Calculating the Magnification

Calculate the magnification using the values found:\[ M = \frac{25.0}{11.76} \approx 2.13 \]
06

Finding the Image Height

Using the magnification and the given object height, find the image height:\[ h_i = M \times h_o = 2.13 \times 1.00 \; \text{mm} \approx 2.13 \; \text{mm} \]The image height is approximately 2.13 mm.

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.

Lens Formula
The lens formula is a fundamental equation used in optics to connect three essential properties of a lens: the object distance (\( u \)), the image distance (\( v \)), and the focal length (\( f \)). This relationship is expressed with the formula:\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]This formula is crucial when using lenses, as it allows us to predict where an image will be formed by determining the position of an object in relation to the lens.In the context of the exercise, if you know the focal length of the lens and either the object or the image distance, you can solve for the unknown distance. When dealing with a thin lens, you can assume that the lens has negligible thickness, simplifying the calculations. This allows for a straightforward application of the lens formula to determine the precise placement of an object or the location of its image.
Focal Length
The focal length is a key characteristic of a lens that defines its ability to bend light and form images. It is the distance between the lens and the point where parallel light rays converge to a single point, known as the focal point. A lens with a shorter focal length bends light more strongly, bringing it to focus at a shorter distance. In our example, the lens has a focal length of 8.00 cm. This property is essential for determining how close or far an object needs to be placed in front of the lens to produce a clear image.
  • A convex lens gathers light to a point after refraction and usually has a positive focal length.
  • A concave lens spreads light apart and typically has a negative focal length.
A clear understanding of focal length is important when using lenses in applications such as cameras and magnifiers to manipulate image formation effectively.
Magnification
Magnification is a measure of how much larger or smaller an image is compared to the actual object. It is represented by the formula:\[M = \frac{h_i}{h_o} = \frac{v}{u}\]Here, \( h_i \) is the image height, \( h_o \) is the object height, \( v \) is the image distance, and \( u \) is the object distance.Magnification quantifies how much a lens can enlarge or reduce the appearance of an object.For instance, if the magnification is greater than 1, the image is larger than the object. If it is less than 1, the image appears smaller.
  • A positive magnification indicates that the image is upright relative to the object.
  • A negative magnification means the image is inverted.
In the exercise example, knowing the object height and using the calculated magnification, you can determine the image height. This concept helps you understand how a lens affects the size perception of an object placed under magnification.

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

(a) Where is the near point of an eye for which a contact lens with a power of \(+2.75\) diopters is prescribed? (b) Where is the far point of an eye for which a contact lens with a power of \(-1.30\) diopters is prescribed for distant vision?

A person can see clearly up close but cannot focus on objects beyond 75.0 \(\mathrm{cm} .\) She opts for contact lenses to correct her vision. (a) Is she nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct her vision? (c) What focal length contact lens is needed, and what is its power in diopters?

One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of 1.55 . When a small leaf is placed 20.0 \(\mathrm{cm}\) in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 \(\mathrm{cm}\) from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33 ) but nothing else were changed?

A converging lens with a focal length of 90.0 \(\mathrm{cm}\) forms an image of a 3.20 -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?

An object is 24.0 \(\mathrm{cm}\) from the center of a silvered spherical glass Christmas tree ornament 6.00 \(\mathrm{cm}\) in diameter. What are the position and magnification of its image?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Kinematics Physics

Read Explanation

Force

Read Explanation

Astrophysics

Read Explanation

Solid State Physics

Read Explanation

Famous Physicists

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.