/*! 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 66 Sketch a ray diagram to show t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sketch a ray diagram to show that if an object is placed less than the focal length from a converging lens, the image is virtual and upright. ( Will tutorial: magnifying glass)

Short Answer

Expert verified
Answer: When an object is placed closer to a converging lens than its focal length, the image formed is virtual and upright. Additionally, the virtual image is larger compared to the object and is located on the same side as the object.

Step by step solution

01

Identify the lens type and focal point

A converging lens is thicker at the middle than at the edges, which causes the light rays to bend toward each other and converge at a point. This point is called the focal point. It is important to notice that a converging lens has two focal points, one on each side of the lens. However, in this problem, we will be considering only the principal focal point that lies on the side of the light rays coming out of the lens.
02

Draw the lens and principal axis

Draw a vertical line representing the converging lens and draw a horizontal line passing through the center of the lens. This horizontal line is called the principal axis.
03

Place the object closer to the lens than its focal length

Place an arrow vertically above the principal axis on the left side of the lens. This arrow represents the object. Make sure to position the object closer to the lens than its focal length, and label this distance as 'd_o', where 'd_o' is less than the focal length 'f'.
04

Draw the three principal rays

From the tip of the object, draw the following three principal rays: 1. Ray 1: Passes through the center of the lens without bending, and continues straight on the other side. 2. Ray 2: Parallel to the principal axis until it reaches the lens, then refracts and passes through the principal focal point on the other side of the lens. 3. Ray 3: Passes through the principal focal point on the object's side of the lens, then refracts to be parallel to the principal axis on the other side of the lens.
05

Notice that the rays do not converge on the right side of the lens

As you can see, the three principal rays diverge after passing through the lens. This means that the image is not formed on the right side of the lens. Instead, we need to trace these rays back behind the lens to find the virtual image.
06

Trace the rays to form the virtual image

Extend the Ray 1 and Ray 2 back behind the lens until they intersect at a point. This point of intersection will be the location of the virtual image. Draw an arrow vertically above the principal axis at this point to represent the virtual image.
07

Observe the image properties

The virtual image formed is located on the same side as the object, and it is larger and upright compared to the object. This confirms that when an object is placed closer to a converging lens than its focal length, the image formed is virtual and upright. You have now successfully sketched a ray diagram to show that when an object is placed closer to a converging lens than its focal length, the resulting image is virtual and upright.

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

A laser beam is traveling through an unknown substance. When it encounters a boundary with air, the angle of reflection is \(25.0^{\circ}\) and the angle of refraction is \(37.0^{\circ} .\) (a) What is the index of refraction of the substance? (b) What is the speed of light in the substance? (c) At what minimum angle of incidence would the light be totally internally reflected?
In order to read his book, Stephen uses a pair of reading glasses. When he holds the book at a distance of \(25 \mathrm{cm}\) from his eyes, the glasses form an upright image a distance of \(52 \mathrm{cm}\) from his eyes. (a) Is this a converging or diverging lens? (b) What is the magnification of the lens? (c) What is the focal length of the lens?
A 5.0 -cm-tall object is placed \(50.0 \mathrm{cm}\) from a lens with focal length \(-20.0 \mathrm{cm} .\) (a) How tall is the image? (b) Is the image upright or inverted?
Apply Huygens's principle to a 5 -cm-long planar wavefront approaching a reflecting wall at normal incidence. The wavelength is \(1 \mathrm{cm}\) and the wall has a wide opening (width \(=4 \mathrm{cm}\) ). The center of the incoming wavefront approaches the center of the opening. Repeat the procedure until you have wavefronts on both sides of the wall. Without worrying about the details of edge effects, what are the general shapes of the wavefronts on each side of the reflecting wall? Repeat Problem 2 for an opening of width $0.5 \mathrm{cm} .$
A point source of light is placed \(10 \mathrm{cm}\) in front of a concave mirror; the reflected rays are parallel. What is the radius of curvature of the mirror?
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.