/*! 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 64 A light bulb is 4.00 \(\mathrm{m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A light bulb is 4.00 \(\mathrm{m}\) from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

Short Answer

Expert verified
Place the mirror 9.00 m from the wall; radius of curvature is 14.4 m.

Step by step solution

01

Understand the Problem

We have a light bulb 4.00 m from a wall. We need to use a concave mirror to project an image on this wall that is 2.25 times the size of the bulb.
02

Determine the Image Distance

Using the magnification formula of mirrors where magnification, \(m = \frac{-v}{u} = 2.25\), and knowing the object distance \(u = 4.00 \text{ m}\), we have \(2.25 = \frac{-v}{4.00}\). Solving for \(v\), we get \(v = -9.00 \text{ m}\). The negative sign indicates the image is real and formed on the opposite side of the object.
03

Calculate the Mirror Distance from the Wall

To find the distance \(d\) of the mirror from the wall, use the fact \(v\), the image distance, is measured from the mirror to the image on the wall, which is 9.00 m. Hence, \(d = 9.00 \text{ m}\).
04

Use the Mirror Formula to Find the Radius of Curvature

The mirror formula is \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\), where \(f\) is the focal length. We have \(\frac{1}{f} = \frac{1}{4.00} + \frac{1}{-9.00}\). Calculating, \(\frac{1}{f} = \frac{5}{36}\), so \(f = 7.20 \text{ m}\). The radius of curvature \(R = 2f = 14.4 \text{ m}\).

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.

Understanding the Mirror Formula
The mirror formula is a fundamental equation in optics, especially when dealing with concave mirrors. It connects the object distance \( u \, \text{m} \,\), the image distance \( v \, \text{m} \,\), and the focal length \( f \, \text{m} \,\) of a mirror through the relationship: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v}. \] This equation helps us calculate one of these distances when the other two are known. In this exercise, with the object distance \( u = 4.00 \, \text{m} \,\) and image distance \( v = -9.00 \, \text{m} \,\), substituting into the formula resolves to the focal length: \[ \frac{1}{f} = \frac{1}{4.00} + \frac{1}{-9.00}. \] Solving gives \( f = 7.20 \, \text{m} \,\). Knowing this helps in finding critical properties of the mirror.
Magnification and Its Importance
Magnification is a measure of how much larger or smaller an image is compared to the object itself. For mirrors, the magnification formula is \( m = \frac{-v}{u} \,\), where \( m \,\) is the magnification, \( v \,\) is the image distance, and \( u \,\) is the object distance. In our exercise, the magnification was given as \( 2.25 \,\), meaning the image is \( 2.25 \,\) times the size of the object. To find \( v \,\), we rearrange the formula: \( 2.25 = \frac{-v}{4.00} \, \Rightarrow \, v = -9.00 \, \text{m} \,\). The negative sign here indicates that the image is real and inverted. This relationship helps determine the correct alignment and placement of the mirror.
Understanding Radius of Curvature
The radius of curvature \( R \,\) of a mirror is an essential concept in optics. It is the radius of the sphere from which the mirror segment is derived. A concave mirror's radius is important because it determines the mirror's focal length. The relationship between the radius of curvature and the focal length is \( R = 2f \,\). In this exercise, we calculated the focal length \( f = 7.20 \, \text{m} \,\), then found the radius of curvature to be \( R = 2 \times 7.20 = 14.4 \, \text{m} \,\). Knowing the radius of curvature is crucial for understanding how a mirror focuses light and forms images. It tells us about the mirror's shape and its optical power.

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 telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the \(95.0-\mathrm{cm}\) lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

Pinhole Camera. A pinhole camera is just a rectangular box with a tiny hole in one face. The film is on the face opposite this hole, and that is where the image is formed. The camera forms an image without a lens, (a) Make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (Hint: Put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) A certain pinhole camera is a box that is 25 \(\mathrm{cm}\) square and 20.0 \(\mathrm{cm}\) deep, with the hole in the middle of one of the \(25 \mathrm{cm} \times 25 \mathrm{cm}\) faces. If this camera is used to photograph a fierce chicken that is 18 \(\mathrm{cm}\) high and 1.5 \(\mathrm{m}\) in front of the camera, how large is the image of this bird on the film? What is the magnification of this camera?

The focal length of the eyepiece of a certain microscope is 18.0 \(\mathrm{mm}\) . The focal length of the objective is 8.00 \(\mathrm{mm}\) . The distance between objective and eyepiece is \(19.7 \mathrm{cm} .\) The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

An object 0.600 \(\mathrm{cm}\) tall is placed 16.5 \(\mathrm{cm}\) to the left of the vertex of a concave spherical mirror having a radius of curvature of \(22.0 \mathrm{cm} .\) (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

A spherical, concave, shaving mirror has a radius of curvature of \(32.0 \mathrm{cm} .\) (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virual? (c) Draw a principal-ray diagram showing the formation of the image.

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.