Physics 1102 Quiz 5G - Exp. 2 April 10- 15, 2017 DIRECTIONS: Work the problem. Y
ID: 1776410 • Letter: P
Question
Physics 1102 Quiz 5G - Exp. 2 April 10- 15, 2017 DIRECTIONS: Work the problem. You must show all work to receive credit, and you must return this sheet Be sure to explain carefully your answers and show your work in a neat and organized manner. The problem is worth 20 points. A student set up a magnetic optical bench with the crossed arrow target (CAT), lens, and screen. Please assume that all positions given here have the offsets already included. The object on the CAT is 1.20 cm high. A lens is placed at 40.0 cm, and an image 0.900 cm high is formed on a screen at 60.0 cm. (a) Calculate the magnitude of the magnification. (b) Calculate the position of the CAT. (c) Calculate the focal length of the lens. (d) Calculate the new position of the lens when a second image is formed, assuming the CAT and screen are not moved. (e) Calculate the size of the image for the case in part (d). -Lau 140Explanation / Answer
a) Magnitude of magnification = image height /object height = 0.9/1.2 = 3/4
b) Magnitude of magnification = distance of image from lens/distance of object from lens
3/4 = 60-40 / (40-x)
position of CAT x = 13.3 cm
c) 1/f = 1/do + 1/di = 1/26.7 + 1/ 20
f = 11.42 cm
d) second image is foremd when for new position of lens, such that object and image distances get interchanged. So now object distance from lens should be 20 cm. Hence position of lens is 13.3 + 20 = 33.3 cm
e) In this case magnification is inverse of magnification in first position, that is 4/3.
size of image = 1.2*(4/3) = 1.6 cm