/*! 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 An object is 16.0 \(\mathrm{cm}\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 38

An object is 16.0 \(\mathrm{cm}\) to the left of a lens. The lens forms an image 36.0 \(\mathrm{cm}\) to the right of the lens. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.00 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

Short Answer

Expert verified
(a) Focal length is -28.8 cm, the lens is diverging. (b) Image height is 18.00 mm, the image is erect. (c) Principal-ray diagram shows diverging rays from the lens.

Step by step solution

01

Identify Known Values

We are given that the object distance \(d_o = -16.0\, \text{cm}\) (negative because it's to the left of the lens) and the image distance \(d_i = 36.0\, \text{cm}\) (positive because it's to the right of the lens). Also, the object height \(h_o = 8.00\, \text{mm}\).
02

Use the Lens Formula to Calculate Focal Length

The lens formula is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Substitute the known values: \(\frac{1}{f} = \frac{1}{-16.0} + \frac{1}{36.0}\). Solving this gives us \(\frac{1}{f} = -0.0625 + 0.0278 = -0.0347\), so \(f \approx -28.8\, \text{cm}\).
03

Determine the Nature of the Lens

Since the focal length \(f\) is negative, the lens is a diverging lens.
04

Calculate the Magnification

The magnification \(m\) is given by \(m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\). Substitute the known values: \(m = -\frac{36.0}{-16.0} = 2.25\).
05

Calculate the Image Height

Using \(h_i = m \times h_o = 2.25 \times 8.00\, \text{mm} = 18.00\, \text{mm}\), the image height is \(18.00\, \text{mm}\).
06

Determine the Orientation of the Image

The magnification \(m = 2.25\) is positive, indicating that the image is erect.
07

Draw a Principal-ray Diagram

To draw the principal-ray diagram for a diverging lens: 1. Draw the lens and principal axis. 2. Draw an incident ray parallel to the principal axis from the top of the object; this ray refracts as if it originated from the focal point on the object's side. 3. Draw an incident ray through the center of the lens, passing straight through without deviation. 4. These rays appear to diverge, but trace backward to form an image on the same side as the object.

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.

Image Distance
When working with lenses, understanding the concept of image distance is crucial. The image distance (\(d_i\)) is the distance from the lens to the point where the image is formed. This can be to the right of the lens or to the left. In this exercise, the image is located 36.0 cm to the right of the lens, making the image distance positive.
Positive image distances indicate that the formed image is on the opposite side from the object (real image). An important aspect to remember is to use the sign conventions appropriately for lenses:
  • If the image is formed on the opposite side to the object, the image distance is positive.
  • If the image is formed on the same side as the object, the image distance is negative (virtual image).
This image distance helps in calculating other properties like focal length and magnification using the lens formula.
Focal Length
The focal length (\(f\)) of a lens is the distance from the lens to the focal point, where light rays parallel to the principal axis converge (for converging lenses) or appear to diverge from (for diverging lenses).
In the lens formula, given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), it aids us in determining whether a lens is converging or diverging based on its sign.
For this problem, substituting the object distance (\(d_o\)) and image distance (\(d_i\)) into the lens formula yields \(f \approx -28.8\, \text{cm}\), indicating a negative focal length.
  • Negative focal lengths mean the lens is diverging.
  • Positive focal lengths denote a converging lens (like magnifying glasses).
The focal length not only tells us the type of lens but also how strongly it converges or diverges rays.
Magnification
Magnification (\(m\)) is a measure of how much larger or smaller an image is compared to the object itself. It is given by the formula: \(m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\). Here, \(h_i\) is the image height, and \(h_o\) is the object height.
In our case, the calculated magnification is 2.25. This positive value tells us two important things:
  • The image size is 2.25 times larger than the object.
  • A positive magnification indicates an erect image.
Negative magnification would imply an inverted image.
By multiplying the object height by magnification (\(h_i = m \times h_o\)), we find the image height to be 18.00 mm. Hence, in this problem, the image is both larger and upright compared to the object.
Principal-ray Diagram
The principal-ray diagram is a helpful tool in visualizing how light interacts with a lens. It shows the paths of specific rays and how they bend or appear to bend when passing through a lens.
For a diverging lens like the one in this problem, several key rays must be drawn:
  • Draw the lens and the principal axis first.
  • An incident ray parallel to the principal axis refracts and appears to originate from the focal point on the same side as the object.
  • Another ray, passing through the center of the lens, goes straight through without deviation.
These ray paths, when traced back, help identify where the virtual image is formed.
For converging lenses, the rays would actually meet on the opposite side, forming a real image. Principal-ray diagrams effectively demonstrate the concepts of lens behavior and image formation visually.

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 speck of dirt is embedded 3.50 \(\mathrm{cm}\) below the surface of a sheet of ice \((n=1.309) .\) What is its apparent depth when viewed at normal incidence?

A person with a near point of \(85 \mathrm{cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 \(\mathrm{cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

A converging lens with a focal length of 12.0 \(\mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

Contact Lenses. 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 \(\mathrm{cm}\) from his eyes instead of the usual 25.0 \(\mathrm{cm} .(\mathrm{a})\) 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 layer of benzene \((n=1.50) 4.20 \mathrm{cm}\) deep floats on water \((n=1.33)\) that is 6.50 \(\mathrm{cm}\) deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Electrostatics

Read Explanation

Physics of Motion

Read Explanation

Translational Dynamics

Read Explanation

Quantum Physics

Read Explanation

Measurements

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.