Explanation Given info: The distance between an object and its upright image is d , the magnification is M . Write the equation for magnification. M = − q p q = − M p (1) Here, M is the magnification. q is the distance between an image and the lens. p is the distance between an object and the lens. Here, the image is given upright. So, for a single lens the object and its image must on the same side of the lens. It means the distance between an object and the lens is negative q < 0 and the magnification M is positive. The distance between an object and its upright image, when the image is in between the object and the lens is, d = p − | q | d = p − ( − q ) ( q < 0 ) d = p + q q = d − p Substitute ( d − p ) for q in the equation (1). ( d − p ) = − M p d p − 1 = − M d p = 1 − M p = d 1 − M Write the thin lens equation. 1 p + 1 q = 1 f (2) Here, f is the focal length of the lens. Substitute − M p for q in equation (2) to get the focal length of the lens. 1 p + 1 ( − M p ) = 1 f M M p − 1 M p = 1 f ( M − 1 ) M p = 1 f f = M p M − 1 (3) Substitute ( d 1 − M ) for p in equation (3) to get the focal length of the lens. f = M ( d 1 − M ) M − 1 = M M − 1 ( d 1 − M ) = − M d ( 1 − M ) 2 Thus, the focal length of the lens when the image is in between the object and the lens is − M d ( 1 − M ) 2

9th Edition

ISBN: 8220100461262

Author: SERWAY

Publisher: Cengage Learning US

See similar textbooks

Videos

Question

Chapter 36, Problem 36.74AP

To determine

The focal length of the lens.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 36, Problem 36.73AP

Chapter 36, Problem 36.75AP

Students have asked these similar questions

A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).

What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V) ammeter I =

simple diagram to illustrate the setup for each law- coulombs law and biot savart law

Chapter 36 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

Ch. 36 - You are standing approximately 2 m away from a...Ch. 36 - You wish to start a fire by reflecting sunlight...Ch. 36 - Consider the image in the mirror in Figure 35.14....Ch. 36 - Prob. 36.4QQ Ch. 36 - Prob. 36.5QQ Ch. 36 - What is the focal length of a pane of window...Ch. 36 - A camera can be modeled as a simple converging...Ch. 36 - Two campers wish to start a fire during the day....Ch. 36 - The faceplate of a diving mask can be ground into...Ch. 36 - Lulu looks at her image in a makeup mirror. lt is...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

The focal length of the lens.

Solution Summary: The author explains how to determine the focal length of the lens. The distance between an object and its upright image is d.

Videos

The focal length of the lens.

Want to see the full answer?

Chapter 36 Solutions