17th Edition

ISBN: 9781938168185

Author: William Moebs, Jeff Sanny

Publisher: OpenStax

Concept explainers

Optics is the study of light in the field of physics. It refers to the study and properties of light. Optical phenomena can be classified into three categories: ray optics, wave optics, and quantum optics. Geometrical optics, also known as ray optics, is…

Converging Lens

Converging lens, also known as a convex lens, is thinner at the upper and lower edges and thicker at the center. The edges are curved outwards. This lens can converge a beam of parallel rays of light that is coming from outside and focus it on a point on…

Plano-Convex Lens

To understand the topic well we will first break down the name of the topic, ‘Plano Convex lens’ into three separate words and look at them individually.

Lateral Magnification

In very simple terms, the same object can be viewed in enlarged versions of itself, which we call magnification. To rephrase, magnification is the ability to enlarge the image of an object without physically altering its dimensions and structure. This pr…

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Textbook Question

Chapter 2, Problem 64P

Two convex lenses of focal lengths 20 cm and 10 cm are placed 30 cm apart, with the lens with the longer focal length on the right. An object of height 2.0 cm is placed midway between them and observed through each lens from the left and from the right. Describe what you will see, such as where the image(s) will appear, whether they will be upright or inverted and their magnifications.

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Chapter 2, Problem 63P

Chapter 2, Problem 65P

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An object of height 2.9 cm is placed at 26 cm in front of a diverging lens of focal length, f = –17 cm. Behind the diverging lens, there is a converging lens of focal length, f = 17 cm. The distance between the lenses is 5 cm. In the next few steps, you will find the location and size of the final image. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is d(i1) = . (Use the sign to indicate which side the image is on; positive sign means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) Where is the final image formed by the second converging lens? Image distance from second lens is d(i2) = (Use the sign to indicate which side the image is on.) How large is the intermediate image formed by the first diverging lens? Intermediate image height is h(i1) = (Use the sign to indicate whether the image is upright (positive) or inverted (negative).) How large is the final image…

- 16 cm. - An object of height 3.2 cm is placed at 22 cm in front of a diverging lens of focal length, f = Behind the diverging lens, there is a converging lens of focal length, f 16 cm. The distance between the lenses is 4 cm. In the next few steps, you will find the location and size of the final image. Hint a. Where is the intermediate image formed by the first diverging lens? Image distance from first lens is cm. (Use the sign to indicate which side the image is on; positive sign dil means image is on the side of outgoing rays, and negative sign means image is on the side opposite to the outgoing rays.) b. Where is the final image formed by the second converging lens? Image distance from second lens is cm. (Use the sign to indicate which side the image is on.) d₁2 = c. How large is the intermediate image formed by the first diverging lens? Intermediate image height is cm. (Use the sign to indicate whether the image is upright (positive) or hil = inverted (negative).) = d. How large…

if its mmage on the fim is to be in focus. What is the magnification? (b) An extension tube is added between the lens and the camera body so that the lens can be positioned 100 mm from film. How close can the object be now? What is the magnification? 5. The focal length of a diverging lens is negative. If ƒ= −20 cm for a particular diverging lens, where will the image be formed of an object located 50 cm to the left of the lens on the optical axis? What is the magnification of the image? 6. The equation connecting s, p, and ffor a simple lens can be employed for spherical mirrors, too. A concave mirror with a focal length of 8 cm forms an image of a small object placed 10 cm in front of the mirror. Where will this image be located? 17. If the mirror described in the previous problem is used to form an image of the same object now located 16 cm in front of the mirror, what would the new image position be? Assuming that the magnification equations developed for

Chapter 2 Solutions

University Physics Volume 3

Ch. 2 - What are the differences between real and virtual...Ch. 2 - Can you see a virtual image? Explain your...Ch. 2 - Can you photograph a virtual image?Ch. 2 - Can you project a virtual image onto a screen?Ch. 2 - Is it necessary to project a real image onto a...Ch. 2 - Devise an arrangement of mirrors allowing you to...Ch. 2 - If you wish to see your entire body in a flat...Ch. 2 - At what distance is an image always located: at...Ch. 2 - Under what circumstances will an image be located...Ch. 2 - What is meant by a negative magnification? What is...

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