/*! 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 65 A concave mirror is to form an i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.00 \(\mathrm{m}\) from the mirror. The filament is 6.00 \(\mathrm{mm}\) tall, and the image is to be 36.0 \(\mathrm{cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?

Short Answer

Expert verified
(a) Place filament 0.1333 m in front of the mirror. (b) Radius of curvature is 0.2624 m.

Step by step solution

01

Understand the problem

We need to find two things: (a) the position of the filament in front of the concave mirror to meet the given conditions, and (b) the radius of curvature of the mirror. We have an image distance (v = 8.00 m), an image height (36.0 cm), and a filament height (6.00 mm). Convert all measurements to consistent units first.
02

Convert Units

Convert the given image height from cm to mm for consistency: 36.0 cm = 360 mm. All distances will remain in meters unless stated otherwise.
03

Use the Magnification Formula

The magnification (M) is given by the formula: \( M = \frac{h_i}{h_o} = \frac{v}{u} \), where \( h_i = 360 \, \mathrm{mm} \) (image height), \( h_o = 6.00 \, \mathrm{mm} \) (object height), \( v = 8.00 \, \mathrm{m} \) (image distance), and \( u \) is the object distance we need to find. First, calculate M: \( M = \frac{360}{6} = 60 \). Hence, \( 60 = \frac{8.00}{u} \). Solve for \( u \).
04

Calculate Object Distance

From Step 3, \( u = \frac{8.00}{60} \). Compute \( u \): \( u = 0.1333 \, \mathrm{m} \) or 133.3 mm. So, the filament should be placed 0.1333 m or 133.3 mm in front of the mirror.
05

Use Mirror Formula to Find Radius of Curvature

The mirror formula relates object distance (u), image distance (v), and focal length (f): \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \). Rearrange for f: Substitute \( u = 0.1333 \, \mathrm{m} \) and \( v = 8.00 \, \mathrm{m} \). Find \( f \).
06

Calculate Focal Length and Use to Find Radius of Curvature

Calculate \( \frac{1}{f} = \frac{1}{0.1333} + \frac{1}{8.00} \), so \( \frac{1}{f} = 7.5 + 0.125 = 7.625 \). This gives \( f = \frac{1}{7.625} = 0.1312 \, \mathrm{m} \). The radius of curvature \( R = 2f \), so \( R = 2 \times 0.1312 = 0.2624 \, \mathrm{m} \) or 262.4 mm.

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.

Optics
Optics is the branch of physics that deals with the study of light and its interactions with different materials. A critical part of optics is understanding how lenses and mirrors affect light.
Concave mirrors are a type of spherical mirror that curves inward, similar to the interior of a bowl. These mirrors can focus light because their shape reflects light rays toward a common point, known as the focal point.
In applications like telescopes, headlights, and cosmetic mirrors, concave mirrors are preferred. Their ability to converge light enhances visibility by creating magnified images or concentrating light on specific areas.
Image Formation
Image formation with concave mirrors involves the reflection of light rays that meet at a point to create a visible image. The nature and position of the image formed depend on the object's distance from the mirror.
  • If the object is placed beyond the center of curvature, the image will be real, inverted, and smaller.
  • At the center of curvature, the image is the same size and inverted.
  • Between the focal point and center, the image becomes larger and remains inverted.
  • Placing it at the focal point results in no image formation, as rays become parallel.
  • Inside the focal point region, the image shifts to virtual, erect, and magnified.
Understanding these principles is key to manipulating the image characteristics using concave mirrors.
Mirror Formula
The mirror formula is a mathematical relation that connects object distance (\(u\)), image distance (\(v\)), and the mirror's focal length (\(f\)). It is expressed as: \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\). This formula is crucial when solving problems involving concave mirrors, as it helps compute unknown quantities like the focal length or object distance.
For our exercise, using the mirror formula allows us to figure out where an object should be positioned concerning a concave mirror to form the desired image at a particular spot. It creates a mathematical framework for anticipating how a mirror will interact with light to produce images.
Magnification
Magnification reveals how much larger or smaller an image is compared to the actual object. It is defined as the ratio of the image height (\(h_i\)) to the object height (\(h_o\)). Mathematically, it's given by the equation: \(M = \frac{h_i}{h_o} = \frac{v}{u}\).
In the context of a concave mirror, magnification tells us if the image will appear smaller or larger than the real object. A magnification value greater than one indicates an enlarged image, while a value less than one indicates reduction.
By leveraging the concept of magnification, we can find distances and sizes when working with mirror problems, as seen in the exercise where these principles are applied to find the proper filament placement and size in the optical system.

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

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 \(\mathrm{cm}\) . Find the position of the image of a small object on the axis, 80.0 \(\mathrm{cm}\) to the left of the first lens.

Choosing a Camera Lens. The picture size on ordinary \(35-\mathrm{mm}\) camera film is \(24 \mathrm{mm} \times 36 \mathrm{mm}\) . Focal lengths of lenses available for \(35-\mathrm{mm}\) cameras typically include \(28,35,50\) (the "normal" lens), \(85,100,135,200,\) and \(300 \mathrm{mm},\) among others. Which of these lenses should be used to photograph the following objects, assuming that the object is to fill most of the picture area? (a) a building 240 \(\mathrm{m}\) tall and 160 \(\mathrm{m}\) wide at a distance of 600 \(\mathrm{m}\) , and (b) a mobile home 9.6 \(\mathrm{m}\) in length at a distance of 40.0 \(\mathrm{m}\) .

Sketch the various possible thin lenses that can be obtained by combining two surfaces whose radii of curvature are 4.00 \(\mathrm{cm}\) and 8.00 \(\mathrm{cm}\) in absolute magnitude. Which are converging and which are diverging? Find the focal length of each if the surfaces are made of glass with index of refraction 1.60 .

The Yerkes refracting telescope of the University of Chicago has an objective 1.02 \(\mathrm{m}\) in diameter with an \(f\) -number of \(19.0 .\) (This is the largest-diameter refracting telescope in the world.) What is its focal length?

One end of a long glass rod is ground to a convex hemispherical shape. This glass an index of refraction of \(1.55 .\) When a small leaf is placed 20.0 \(\mathrm{cm}\) in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 \(\mathrm{cm}\) from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33\()\) but nothing else were changed?

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.