/*! 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 29 An insect \(3.75 \mathrm{~mm}\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 29

An insect \(3.75 \mathrm{~mm}\) tall is placed \(22.5 \mathrm{~cm}\) 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 \(13.0 \mathrm{~cm},\) 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.

Short Answer

Expert verified
Part (a): The calculated focal length will indicate properties of the lens. The image position obtained will tell whether it's real or virtual. The magnification and height will reveal the size and orientation. Part (b): Similar steps with new values after lens reversal.

Step by step solution

01

Finding the focal length of the lens

The lens maker's formula is given by \[\frac{1}{f} = (n-1) \left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right)\] Here, \(n=1.70, R_{1}=\infty\) (since left surface is flat), and \(R_{2}=-13.0\) cm (negative since we are considering the curvature as seen from the object point). Using these values, calculate the focal length \(f\).
02

Determine the position of the image

Use the lens formula given by \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] Here, \(u=-22.5\) cm (negative since the object is on the same side of the light entering the lens). Substitute the known values to find the image distance \(v\). If \(v\) is positive, the image is formed on the opposite side of the light entering the lens (real image); if it's negative, it's on the same side (virtual image).
03

Determine the nature and size of the image

The magnification \(m\) is given by the formula \[m=-\frac{v}{u}\] Substitute the known values to get \(m\). If \(m>0\), the image is erected and if \(m<0\), it is inverted. The size of the image \(h'\) is given by \(h' = m \times h\) where \(h=3.75\) mm is the height of the object.
04

Repeat the process for reversed lens

When the lens is reversed, the radius of curvature \(R_{1}=-13.0\) cm and \(R_{2}=\infty\). Repeat steps 1-3 using these values to find the new focal length, image distance, image size and nature.

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 Maker's Formula
The Lens Maker's Formula is crucial for understanding how lenses focus light to form an image. This formula helps us calculate the focal length of a lens, which determines how strongly it converges or diverges light. The formula is given by: \[ \frac{1}{f} = (n-1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right) \] Here:
  • \(f\) is the focal length of the lens.
  • \(n\) is the index of refraction of the lens material.
  • \(R_{1}\) and \(R_{2}\) are the radii of curvature of the two lens surfaces.
It鈥檚 important to remember that if one side of the lens is flat, like in a planoconvex lens, its radius of curvature is considered infinite \(\(R_{1} = \infty\)\). This simplifies the computation, making it easier to determine the focal length.
Image Formation
In geometric optics, image formation involves determining where and how an object is projected by a lens. The position and nature of the image depend heavily on the lens鈥檚 properties and its orientation relative to the object.
To find the image location, we use the lens formula:\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]Where:
  • \(u\) is the object distance, which is negative as per our sign convention since the object is placed on the side from which the light enters the lens.
  • \(v\) is the image distance. The sign of \(v\) tells us about the nature of the image: a positive \(v\) means the image is on the opposite side of the lens (real), whereas a negative \(v\) indicates it鈥檚 on the same side as the object (virtual).
By correctly applying this formula, you can easily locate where the image forms.
Focal Length Calculation
Focal length calculation is at the heart of understanding how lenses behave. By using Lens Maker鈥檚 Formula, one can derive the focal length, which is pivotal in determining how a lens focuses light.For our specific case:- The radius of curvature \(R_{1}\) is infinity because one side of the lens is flat.- The other side \(R_{2}\) has a radius of \(-13.0\) cm. The index of refraction \(n\) is given as 1.70. Substituting these values into the Lens Maker's Formula determines the focal length of this planoconvex lens. Accurately obtaining the focal length is essential for further analysis or when the lens鈥檚 configuration changes.
Magnification
Magnification demonstrates how much larger or smaller an image is compared to the object itself. It's described by the formula:\[ m = -\frac{v}{u} \]In this formula:
  • Negative sign indicates inversion of the image. Thus, if magnification \(m\) has a positive value, the image is upright. Conversely, a negative \(m\) implies an inverted image.
  • The magnitude of \(m\) defines the image size compared to the object.
To find the actual height of the image \(h'\), you can use:\[ h' = m \times h \]Where \(h\) is the initial height of the object (3.75 mm in our case). By calculating \(m\) and \(h'\), we acquire insights into the dimensions and orientation of the image formed by the lens.
Curved Lenses
Curved lenses, such as planoconvex lenses, are widely used in optics to manipulate light paths. Understanding their structure and function is essential in many scientific and practical fields. In a planoconvex lens:
  • One surface is flat while the other is curved, providing different focusing properties depending on which side faces the object or source of light.
  • When light enters through the curved side first, it bends more strongly than when the flat side faces the incoming light.
Each lens orientation affects the path of light, changing where and how the image is formed. Thus, understanding the lens geometry and material allows for precise image control vital for cameras, glasses, and instruments relying on precise imaging.

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 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?

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere \(2.50 \mathrm{~cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells \(5.0 \mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of \(25 \mathrm{~cm} ?\) (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{~min}\right),\) and compare it with the typical experimental value of about \(1.0 \mathrm{~min} .\) (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

To determine the focal length \(f\) of a converging thin lens, you place a \(4.00-\mathrm{mm}\) -tall object a distance \(s\) to the left of the lens and measure the height \(h^{\prime}\) of the real image that is formed to the right of the lens. You repeat this process for several values of \(s\) that produce a real image. After graphing your results as \(1 / h^{\prime}\) versus \(s\), both in \(\mathrm{cm}\), you find that they lie close to a straight line that has slope \(0.208 \mathrm{~cm}^{-2}\). What is the focal length of the lens?

Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius of curvature of \(15.0 \mathrm{~cm} .\) The refractive index of the glass is \(1.50 .\) What is the distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod? Is the image in the air or in the glass?

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 \(M ?\)
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Energy Physics

Read Explanation

Electromagnetism

Read Explanation

Fields in Physics

Read Explanation

Modern Physics

Read Explanation

Magnetism and Electromagnetic Induction

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.