/*! 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 95 Three thin lenses, each with a f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 95

Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens.

Short Answer

Expert verified
The final image position is calculated using the lens formula for each lens, resulting in a stepwise determination of image positions for the second and third lenses.

Step by step solution

01

Understand the Lens Formula

To solve this problem, use the lens formula, which 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 from the lens. Here, \( f = 40.0 \) cm.
02

Calculate Image Position for the First Lens

Given that the object distance \( d_o = -80.0 \) cm (negative as it is to the left of the lens), use the lens formula: \( \frac{1}{40} = \frac{1}{-80} + \frac{1}{d_{i1}} \). Solve for \( d_{i1} \):\[ \frac{1}{d_{i1}} = \frac{1}{40} + \frac{1}{80} = \frac{3}{80} \Rightarrow d_{i1} = \frac{80}{3} \text{ cm (positive, to the right of the first lens)} \].
03

Calculate Object Position for the Second Lens

Find the object distance for the second lens: The image from the first lens becomes the virtual object for the second lens. The position of this object relative to the second lens is \( d_{o2} = 52.0 - \frac{80}{3} = \frac{76}{3} \text{ cm} \).
04

Calculate Image Position for the Second Lens

Use the lens formula for the second lens: \( \frac{1}{40} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}} \), where \( d_{o2} = \frac{76}{3} \). Solve for \( d_{i2} \):\[ \frac{1}{d_{i2}} = \frac{1}{40} - \frac{3}{76} \Rightarrow d_{i2} = \text{calculate and find the value} \].
05

Calculate Object Position for the Third Lens

Find the object distance for the third lens: The image from the second lens becomes the virtual object for the third lens. Position it 52.0 cm ahead of the third lens.
06

Calculate Final Image Position

Repeat the lens formula for the third lens with the object distance found in Step 5. Calculate \( d_{i3} \) to find the final image position after passing through all three lenses.

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 in geometric optics that relates the focal length (\( f \)), the image distance (\( d_i \)), and the object distance (\( d_o \)). It's expressed as:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
This equation allows us to calculate any one of these distances if the other two are known.
In the context of lenses, the focal length is the distance from the lens to the principal focus, where parallel rays converge.
The object distance is the distance from the object to the lens, while the image distance is the distance from the lens to the projected image.
Understanding and manipulating this formula is essential for solving problems involving thin lenses, like in this exercise.
Focal Length
Focal length (\( f \)) is an intrinsic property of a lens, defined as the distance over which initially collimated rays are brought to a focus. A positive focal length indicates a converging lens, while a negative one indicates a diverging lens.
In this exercise, all lenses have the same focal length of 40.0 cm, which is typical for converging lenses.
  • Shorter focal lengths: lenses that are more powerful in bending light rays.
  • Longer focal lengths: lenses that bend light rays less strongly.
In any optical setup, knowing the focal length is crucial because it affects how the image is formed and its characteristics, such as size and orientation.
Image Distance
Image distance (\( d_i \)) is the measure of how far from the lens the image is formed.
In our example, we calculate the image position relative to each lens, moving from one to the next.
By using the lens formula:
  • If \( d_i \) is positive: the image is on the opposite side of the lens from the object (real image).
  • If \( d_i \) is negative: the image is on the same side as the object (virtual image).
Each lens affects the subsequent image distance because the image from one lens serves as the object for the next.
This step-by-step approach is vital to solve multi-lens systems.
Object Distance
Object distance (\( d_o \)) is the distance from the object to the lens and can be positive or negative based on its position. For concave systems like in this exercise, it starts negative as it's left of the lens.
Considerations include:
  • \( d_o \) is negative when the object is on the same side as incoming light (real object).
  • It becomes positive if the object is a reflected image projected on the other side of the lens (virtual object).
Managing object distances is crucial for cascading lens systems, as the image from one becomes the next lens's object. Properly calculating these distances ensures accurate image placements inside a multiple lens setup.

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

An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A converging lens with a focal length of 70.0 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 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 placed 22.0 cm from a screen. (a) At what two points between object and screen may a converging lens with a 3.00-cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Electricity

Read Explanation

Electrostatics

Read Explanation

Physics of Motion

Read Explanation

Energy Physics

Read Explanation

Oscillations

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.