/*! 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} Q45PE A camera lens used for taking cl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is 33.0 mm.

(a) What is the closest object that can be photographed?

(b) What is the magnification of this closest object?

Short Answer

Expert verified

(a) The distance for the closest object to be photographed is d∘=66.0mm.

(b) The magnification of the closest object is m=-12.

Step by step solution

01

Concept Introduction

The focal length is the distance between a convex lens or a concave mirror and the focal point of a lens or mirror. It is the point at which two parallel light beams meet or converge. Depending on the lens and mirror (concave or convex), the focal length varies with the sign (positive or negative).

02

Information Provided

  • The focal length of the camera lens:22.0mm=221000=22×10-3m.
  • Distance between film and lens: 33.0mm=331000=33×10-3m.
03

Closest Object to be photographed

(a)

Use the thin lens equation for the focal length:

1f=1di+1d∘.

Rearranging the expression and substituting the values –

1d∘=1f-1did∘=122.0×10-3m-133.0×10-3m-1=0.066m=0.066×1000mm=66.0mm

Therefore, the value for distance is obtained as d∘=66.0mm.

04

Calculation for magnification

(b)

Use the equation of the magnification that relates the image distance to the object distance: .

Substituting the values and evaluating –

m=-did∘=-0.033m0.066m=-12

Therefore, the value for magnification is obtained as m=-12.

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!

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

Consider a \(250{\rm{ }}W\) heat lamp fixed to the ceiling in a bathroom. If the filament in one light burns out then the remaining three still work. Construct a problem in which you determine the resistance of each filament in order to obtain a certain intensity projected on the bathroom floor. The ceiling is \(3.0{\rm{ }}m\) high. The problem will need to involve concave mirrors behind the filaments. Your instructor may wish to guide you on the level of complexity to consider in the electrical components.

(a) Using information in Figure \(25.23\), find the height of the instructor’s head above the water, noting that you will first have to calculate the angle of incidence. (b) Find the apparent depth of the diver’s head below water as seen by the instructor.

A narrow beam of light containing red (\(660{\rm{ }}nm\)) and blue (\(470{\rm{ }}nm\)) wavelengths travels from air through a\(1.00{\rm{ }}cm\)thick flat piece of crown glass and back to air again. The beam strikes at a\({30.0^ \circ }\)incident angle. (a) At what angles do the two colors emerge? (b) By what distance are the red and blue separated when they emerge?

Show that when light reflects from two mirrors that meet each other at a right angle, the outgoing ray is parallel to the incoming ray, as illustrated in the following figure.

A scuba diver training in a pool looks at his instructor as shown in Figure 25.53. What angle does the ray from the instructor’s face make with the perpendicular to the water at the point where the ray enters? The angle between the ray in the water and the perpendicular to the water is 25.0°.

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.