Question

Problem 25.25 - Enhanced - with Solution 19 of 21 A Review Constants Submit Previous Answers A thin lens with a focal length of 6.15 cm is used as a simple magnifier. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A simple magnifier. Correct Set Up: When the object is at the focal point the image is at infinity and M= 25.0 cm Solve: M 25.0 cm 6.15cm 4.07x Part B When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at infinity and that the lens is very close to the eye. Enter the smallest distance the object can be from the lens in centimeters. VASO ? d = 4.94 cm Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining

Answer 1

25-25

Part B. 6.15 cm

25-17

Part C :f= -0.71 m.

Part D : Power = -1.408 Dioptre

Part A.

s = 4.699 cm to the left

s' = 14.301 cm to the left.

25-25 6) S - D (giren) f': 6.15 given is at infinity image viewed by eye Ś + Ś ㅗ f' tas tos al at -O - - ar ㅗ 6.15 ㅗ 6.15 st S = 6.15 cm 25.4 → th 78 S = -75cm . S: 2 Ś + = $ + - 78 f = -71 cm H: -0.71 M Power : c $ 1.408 dioptre 5.0.71 ma - - S': -3.043 S Lateral magnification -S' y! 1.40cm S - 3.043 9 0.46 cm S: object distance LO Ś' - image distance y'= height of image y : object height of from i NOW, - 2.043 &+ + 2 = 17 / + - 3043 + _ 3:0436 – 3.0435 2.043 70 3.0435. S= 4.699 cm. s': 4.699 X-3.043 - - 14.301 cm. the object is 4.699 cm to the left The image is 114.301 cm to the left