/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
The final image will be located a distance \(d_3\) cm to the right of the third lens.

Step by step solution

01

Calculate the image position for the first lens

Use the thin lens formula \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i} \), where \(f = 40.0cm\) is the focal length, \(d_0= -80.0cm\) is the object distance (negative since the object is to the left), and \(d_i\) is the image distance. Solve for \(d_i\), rearranging gives \(d_i = \frac{1}{\frac{1}{f} - \frac{1}{d_0}}\).
02

Using the image distance for the first lens, calculate the object distance for the next lens

The second lens object distance will be the image distance from the first lens subtracted from the lense separation, \(d_0^{'} = d_1-52.0 cm\).
03

Calculate the image position for the second lens

Repeat the thin lens formula from step 1 with the object distance from step 2, \(d_= = \frac{1}{\frac{1}{f} - \frac{1}{d_0^{'}}}\).
04

Calculate the object distance for the third lens

The third lens object distance will be the image distance from the second lens subtracted from the lense separation, \(d_0^{''} = d_2-52.0 cm\).
05

Calculate the image position for the third lens

Apply the thin lens formula from step 1 with the object distance from step 4, \(d_3 = \frac{1}{\frac{1}{f} - \frac{1}{d_0^{''}}}\). This gives the final image position.

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 91Ó°ÊÓ!

Key Concepts

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

Focal Length
The focal length is a fundamental concept in optics, describing the distance between the center of a lens and its focal point. This is the point where parallel rays of light converge after passing through the lens. In the exercise, each of the three lenses has a focal length of \(40.0 \text{ cm}\).

Focal length is typically represented by the letter \(f\) and is measured in centimeters or meters, depending on the scale of the problem.
  • A positive focal length indicates a converging lens, which brings light rays together.
  • A negative focal length would imply a diverging lens, spreading light rays apart.
Understanding the role of focal length helps predict how light will behave when interacting with lenses, influencing how images are formed. Being familiar with this concept is critical for effectively using the thin lens equation, \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i} \), to calculate image distances.
Image Distance
Image distance is the measurement from a lens to the location where an image is produced. In mathematical terms, it is represented as \(d_i\) in the thin lens equation. Calculating image distance is crucial for determining where an image forms in relation to a lens.

In our exercise, we calculate the image distance for each of the three lenses separately, using the formula \( d_i = \frac{1}{\left( \frac{1}{f} - \frac{1}{d_0} \right)} \).
  • Image distance is positive when the image forms on the opposite side of the incoming light (real image).
  • It is negative when the image forms on the same side as the object (virtual image).
Knowing whether an image is real or virtual helps understand the properties of the image, such as its orientation and magnification.
Lens Separation
Lens separation is the physical distance between adjacent lenses in a system. In our problem, the distance between each pair of lenses is \(52.0 \text{ cm}\).

This measurement is important as it affects how the image formed by one lens becomes the object for the next lens in the system.
  • For the second lens, the object is the image formed by the first lens minus the separation distance.
  • Similarly, for the third lens, the object distance uses the image from the second lens minus the separation distance.
Understanding lens separation is crucial for systems with multiple lenses, as it impacts the cumulative effect and overall positioning of the images in such a compound lens setup.
Object Distance
Object distance refers to the space between a lens and the object it is focusing on. Often denoted by \(d_0\), this measurement is integral in calculating the position and nature of the resulting image.

In lens calculations, object distance is positive if the object is on the incoming side of the light rays and negative if it is on the same side as the image. In the problem, the object is \(80.0 \text{ cm}\) to the left of the first lens, giving us an initial object distance of \(-80.0 \text{ cm}\).
  • Subsequent object distances depend on the image positions calculated earlier in the lens system.
  • These become crucial inputs for determining image distances for multi-lens problems using the thin lens equation.
A firm comprehension of object distance helps in systematically approaching multi-step lens problems.

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

The focal points of a thin diverging lens are \(25.0 \mathrm{~cm}\) from the center of the lens. An object is placed to the left of the lens, and the lens forms an image of the object that is \(18.0 \mathrm{~cm}\) from the lens. (a) Is the image to the left or right of the lens? (b) How far is the object from the center of the lens? (c) Is the height of the image less than, greater than, or the same as the height of the object?

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?

You are in your car driving on a highway at \(25 \mathrm{~m} / \mathrm{s}\) when you glance in the passenger-side mirror (a convex mirror with radius of curvature \(150 \mathrm{~cm}\) ) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of \(1.9 \mathrm{~m} / \mathrm{s}\) when the truck is \(2.0 \mathrm{~m}\) from the mirror, what is the speed of the truck relative to the highway?

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 \(1.00 \mathrm{~mm}\) high, what is the height of its image formed by the magnifier?

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.

See all solutions

Recommended explanations on Physics Textbooks

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.