#3 in image Convex LensesWhen an object is placed at a distance d, from a lens of focal length f, and d, isgreater than f, the rays hitting the lens will form a real image on the other side, at adistance d; from the lens. The object distance do, image distance d;, and focal length fare related by the "Thin Lens Equation":11+d." diThe height of the image h; is usually different than the height of the originalobject ho. The factor by which the image is bigger than the object is called theMagnification M of the image. The magnification depends on the distances do and d;if the image is farther from the lens than the object, it will be bigger than the object.hịM =dihodoIf the magnification is positive, the image is "right side up"; if it is negative, the imageis upside down or "inverted."In this exercise, you will use the above equations to see what happens to theimage as you move an object closer and closer to the lens. We are looking forpatterns, to see how the variables change in relation to each other.Directions: Here are some practice problems to make you familiar with these concepts.Please review the following.An object is placed at four different distances from a convex lens withf = 20.0 cm:a) do = 60.0 cmb) do = 50.0 cmc) do = 40.0 cmd) do = 30.0 cm1. For each distance, calculate the image distance d;.2. For each distance, calculate the magnification M.3. From your results, what happens to the image distance d; as the object gets closerand closer to F?

focal length of convex lens (f) = 20 cm

An object is placed at four different distances from a convex lens with = 20.0 cm: a) d. = 60.0 cm b) do = 50.0 cm c) do = 40.0 cm d) do = 30.0 cm %3D 1. For each distance, calculate the image distance d;. 2. For each distance, calculate the magnification M. 8. From your results, what happens to the image distance d; as the object gets closer and closer to F?

An object is placed at four different distances from a convex lens with = 20.0 cm: a) d. = 60.0 cm b) do = 50.0 cm c) do = 40.0 cm d) do = 30.0 cm %3D 1. For each distance, calculate the image distance d;. 2. For each distance, calculate the magnification M. 8. From your results, what happens to the image distance d; as the object gets closer and closer to F?

University Physics Volume 3

17th Edition

ISBN:9781938168185

Author:William Moebs, Jeff Sanny

Publisher:William Moebs, Jeff Sanny

Chapter2: Geometric Optics And Image Formation

Section: Chapter Questions

Problem 106P: Construct Your Own Problem Consider a telescope of the type used by Galileo, having a convex...

See similar textbooks

Similar questions

A nearsighted person cannot see objects clearly beyond 25.0 cm (her far point). If she has no astigmatism and contact lenses are prescribed for her, what (a) power and (b) type of lens are required to correct her vision?
What happens to the image produced by a lens if the lens is “stopped down” (the aperture reduced, thereby reducing the amount of light passing through the lens) with an iris diaphragm-a device that covers its periphery?
A diverging lens has a focal length of 20 cm. What is the power of the lens in diopters?
A 7.5 binocular produces an angular magnification of —7.50, acting like a telescope. (Mirrors are used to make the image upright.) If the binoculars have objective lenses with a 75.0-cm focal length, what is the focal length of the eyepiece lenses?
. (a) In a camera equipped with a 50-mm focal-length lens, the maximum distance that the lens can be from the film is 60 mm. What is the smallest distance an object can be from the camera if its image on the film 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?
You can argue that a that piece of glass, such as in a window, is like a lens with an infinite focal length. If so, where does it form an image? That is, how are diand dorelated?
The accommodation limits for a nearsighted persons eyes are 18.0 cm and 80.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly?
The power for normal distant vision is 50.0 D. A severely myopic patient has a far point of 5.00 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him?
Can an image be larger than the object even though its magnification is negative? Explain.
Construct Your Own Problem Consider a telescope of the type used by Galileo, having a convex objective and a concave eyepiece as illustrated in part (a) of Figure 2.40. Construct a problem in which you calculate the location and size of the image produced. Among the things to be considered are the focal lengths of the lenses and their relative placements as well as the size and location of the object. Verify that the angular magnification is greater than one. That is, the angle subtended at the eye by the image is greater than the angle subtended by the object.
A spy satellite orbits Earth at a height of 180 km. What is the minimum diameter of the objective lens in a telescope that must be used to resolve columns of troops marching 2.0 m apart? Assume =550 nm.
Suppose a certain person’s visual acuity is such that he can see objects clearly that form an image 4.00mhigh on his retina. What is the maximum distance at which he can read the 75.0-cm-high letters on the side of an airplane?