/*! 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 105 An object is located 30.0 cm to ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 26: Problem 105

An object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm. (a) Draw a ray diagram to scale and from it determine the image distance and the magnification. (b) Use the thin-lens and magnification equations to verify your answers to part (a).

Short Answer

Expert verified
The image distance is -75.0 cm, and the magnification is 2.5, indicating a virtual and upright image.

Step by step solution

01

Understanding the Problem

We are given a converging lens with a focal length of 50.0 cm and an object placed 30.0 cm to the left of the lens. We need to determine the image distance and magnification both graphically with a ray diagram and analytically using lens equations.
02

Drawing the Ray Diagram

Draw a horizontal line as the optical axis. Place the lens in the center and mark the focal points (F) 50.0 cm on each side of the lens. Position the object 30.0 cm to the left of the lens. Draw three principal rays: 1. A ray parallel to the principal axis refracting through the focal point on the opposite side. 2. A ray passing through the center of the lens which does not bend. 3. A ray passing through the focal point on the object's side refracting parallel to the principal axis. Locate where these rays converge to find the image location.
03

Using the Thin-Lens Equation

The thin-lens equation 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. Substitute \( f = 50.0 \) cm and \( d_o = 30.0 \) cm into the equation to solve for \( d_i \): \( \frac{1}{50} = \frac{1}{30} + \frac{1}{d_i} \). Solve for \( d_i \) to find the image distance.
04

Solving the Image Distance Equation

Rearrange the equation: \( \frac{1}{d_i} = \frac{1}{50} - \frac{1}{30} = \frac{3}{150} - \frac{5}{150} = -\frac{2}{150} \). So, \( d_i = -75.0 \) cm. This indicates the image is located 75.0 cm on the same side as the object, meaning it's virtual.
05

Calculating Magnification

The magnification equation is: \( m = - \frac{d_i}{d_o} \). With \( d_i = -75.0 \) cm and \( d_o = 30.0 \) cm, \( m = - \frac{-75}{30} = 2.5 \). The image is magnified 2.5 times larger and upright since the magnification is positive.

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.

Ray Diagram
When working with lenses, ray diagrams serve as a crucial tool. A ray diagram helps visualize how light travels through a lens to form an image. To draw a ray diagram for a converging lens, follow these steps:

  • Start by drawing a horizontal line, known as the optical axis.
  • Place the lens at the center, marking the focal points (F) on both sides of the lens, in this case, 50 cm away.
  • Position the object on the left side of the lens, 30 cm away.
  • Draw three principal rays from the object:
    • Ray 1: A ray parallel to the optical axis refracts through the focal point on the opposite side.
    • Ray 2: A ray passing through the center of the lens continues straight without bending.
    • Ray 3: A ray traveling through the focal point on the object's side refracts and travels parallel to the optical axis afterwards.
These rays will converge on the object鈥檚 side of the lens, indicating a virtual image location.
Thin-Lens Equation
The thin-lens equation is a mathematical representation used to calculate the image distance formed by a lens with a given focal length. The equation is expressed as follows:

\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]

Where:
  • \( f \) = focal length of the lens (50 cm in this case).
  • \( d_o \) = distance from the object to the lens (30 cm for this example).
  • \( d_i \) = distance from the image to the lens, which we need to find.
To solve for \( d_i \), rearrange the equation:

\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}\]

Applying these values, the calculation reveals that the image distance \( d_i = -75.0 \) cm, representing a virtual image located on the same side as the object.
Image Distance
The image distance, denoted as \( d_i \), reflects where the image is formed relative to the lens. In the context of this exercise, using the thin-lens equation, we determined that \( d_i \) equals -75.0 cm.

Here is what this value signifies:
  • The negative sign indicates the image is virtual, as it forms on the same side of the lens as the object.
  • Virtual images cannot be projected onto a screen since they're formed by the perceived divergence of rays.
Understanding the image distance is crucial in optics, as it gives insight into the image's position and nature, helping you verify your predictions from ray diagrams.
Magnification
Magnification involves determining how much larger or smaller the image appears compared to the object itself, and it is calculated using the magnification equation:

\[m = - \frac{d_i}{d_o}\]

With this equation, we can determine:
  • \( d_i \) = -75.0 cm (image distance).
  • \( d_o \) = 30.0 cm (object distance).
Substituting these values:

\[m = -\left( \frac{-75}{30} \right) = 2.5\]

This result shows:
  • The image is magnified, as it's 2.5 times larger than the actual object size.
  • The positive magnification confirms the image is upright, which also aligns with the virtual image status.
Understanding magnification is key in assessing how lenses alter the perception of object sizes in practical applications.

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 drawing shows a rectangular block of glass \((n=1.52)\) surrounded by liquid carbon disulfide \((n=1.63) .\) A ray of light is incident on the glass at point A with a \(30.0^{\circ}\) angle of inci- dence. At what angle of refraction does the ray leave the glass at point \(\mathrm{B} ?\)

The near point of a naked eye is 32 cm. When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of 16 cm, and is held next to the eye. The enlarged image that is seen is located 64 cm from the magnifying glass. Determine the angular size of the image.

An object is placed 20.0 \(\mathrm{cm}\) to the left of a diverging lens \((f=-8.00 \mathrm{cm})\) . A concave mirror \((f=12.0 \mathrm{cm})\) is placed 30.0 \(\mathrm{cm}\) to the right of the lens. (a) Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?

An object is located 9.0 \(\mathrm{cm}\) in front of a converging lens \((f=6.0 \mathrm{cm}) .\) Using an accurately drawn ray diagram, determine where the image is located.

A dentist is examining a dental filling in a patient鈥檚 tooth. The diameter of the filling is 2.4 mm, and the dentist鈥檚 near point is 17.0 cm. To get a better look at the filling, the dentist dons safety goggles fitted with magnifying glasses \((f=6.0 \mathrm{cm}) .\) Find the greatest possible angular size (in radians) of the patient's filling when viewed by the dentist, both (a) without and \(\quad\) (b) with the magnifying glasses.
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Particle Model of Matter

Read Explanation

Mechanics and Materials

Read Explanation

Magnetism

Read Explanation

Torque and Rotational Motion

Read Explanation

Fluids

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.