Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 63

An object of height \(3.00 \mathrm{cm}\) is placed \(12.0 \mathrm{cm}\) from a diverging lens of focal length \(-12.0 \mathrm{cm} .\) Draw a ray diagram to find the height and position of the image.

Short Answer

Expert verified
Question: An object with a height of 3.00 cm is placed 12.0 cm from a diverging lens with a focal length of -12.0 cm. Find the height and position of the image formed by the lens. Answer: The image is a virtual image located 4.00 cm to the left of the lens and has a height of 1.00 cm.

Step by step solution

01

Identify the object's properties

The object has a height of \(3.00 \mathrm{cm}\) and is placed \(12.0 \mathrm{cm}\) from the diverging lens. The lens has a focal length of \(-12.0 \mathrm{cm}\).
02

Find the focal point of the lens

A diverging lens has a negative focal length, so the focal point is on the same side as the object. In this case, the focal point is \(12 \mathrm{cm}\) to the left of the lens.
03

Draw the ray diagram

Follow these rules to draw the ray diagram: 1. Draw a straight line from the object's top point, parallel to the principal axis, until it reaches the lens. After passing through the lens, this ray will diverge as if it's coming from the focal point. 2. Draw another straight line from the top point of the object through the center of the lens. This ray will continue undeviated. 3. The image is formed where these two rays appear to intersect.
04

Use the lens formula to find the image distance

The lens formula is: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) We know the focal length \(f = -12.0 \mathrm{cm}\) and the object distance \(d_o = 12.0 \mathrm{cm}\). So we can solve for the image distance \(d_i\): \(\frac{1}{-12} = \frac{1}{12} + \frac{1}{d_i} \Rightarrow d_i = -4 \mathrm{cm}\) The negative sign means that the image is virtual and located on the same side of the lens as the object. The image is \(4 \mathrm{cm}\) to the left of the lens.
05

Calculate the image height using the magnification formula

The magnification formula is: \(m = \frac{h_i}{h_o} = \frac{d_i}{d_o}\) We know the object's height \(h_o = 3.00 \mathrm{cm}\), the object distance \(d_o = 12.0 \mathrm{cm}\), and the image distance \(d_i = -4 \mathrm{cm}\). We can now find the image height \(h_i\): \(-\frac{h_i}{3} = \frac{-4}{12} \Rightarrow h_i = 1.00 \mathrm{cm}\) The image height is \(1.00 \mathrm{cm}\). Since the magnification is negative, the image is inverted. So, the image is a virtual image located \(4.00 \mathrm{cm}\) to the left of the lens and has a height of \(1.00 \mathrm{cm}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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) Sunlight reflected from the still surface of a lake is totally polarized when the incident light is at what angle with respect to the horizontal? (b) In what direction is the reflected light polarized? (c) Is any light incident at this angle transmitted into the water? If so, at what angle below the horizontal does the transmitted light travel?
Light travels in a medium with index \(n_{1}\) toward a boundary with another material of index \(n_{2}<n_{1},\) (a) Which is larger, the critical angle or Brewster's angle? Does the answer depend on the values of \(n_{1}\) and \(n_{2}\) (other than assuming \(n_{2}<n_{1}\) )? (b) What can you say about the critical angle and Brewster's angle for light coming the other way (from the medium with index \(n_{2}\) toward the medium with \(n_{1}\) )?
A penny is at the bottom of a bowl full of water. When you look at the water surface from the side, with your eyes at the water level, the penny appears to be just barely under the surface and a horizontal distance of $3.0 \mathrm{cm}\( from the edge of the bowl. If the penny is actually \)8.0 \mathrm{cm}$ below the water surface, what is the horizontal distance between the penny and the edge of the bowl? [Hint: The rays you see pass from water to air with refraction angles close to \(\left.90^{\circ} .\right]\)
A \(45^{\circ}\) prism has an index of refraction of \(1.6 .\) Light is normally incident on the left side of the prism. Does light exit the back of the prism (for example, at point \(P\) )? If so, what is the angle of refraction with respect to the normal at point \(P ?\) If not, what happens to the light?
In her job as a dental hygienist, Kathryn uses a concave mirror to see the back of her patient's teeth. When the mirror is \(1.20 \mathrm{cm}\) from a tooth, the image is upright and 3.00 times as large as the tooth. What are the focal length and radius of curvature of the mirror?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Modern Physics

Read Explanation

Electricity

Read Explanation

Quantum Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Oscillations

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.

Sign-up for free