mon/Take TutorialAssignment.aspx Home I Student: anjall ga 2018 11:59:00 PM End
ID: 2033639 • Letter: M
Question
mon/Take TutorialAssignment.aspx Home I Student: anjall ga 2018 11:59:00 PM End Date: 4/6/2018 4:00.00 AM in front of a convex mirror with radins of curvature IS cm. Homework 6 Begin Date: 3/27/2018 2:00:00 PM-Due Date: 4/S (8%) Problem 12: A 29-cm tall obyect is placed 5.0 st 33% Part (a) what is the image distance, in centimeters? Include its sign Grade Sum Potential Submissions ed cotano i astad : acoeOA ?.tia 61 atan) acctano sinho Attlempts rema 5% per attem detaiied view ed Degrees .) Radars ed ted ted ted 33% Part (b) what is the image height. cettameters? Inchae as sagr. 33% Part (c) What is the orientation of de mage reatte to de dtedr ted OOLBY HONE THEATER Fe F7 F8 F9 F10 F11 F12 Prtsc Pause SysRq Break Ins DefExplanation / Answer
given
h = 2.9 cm
u = 5.4 cm (object distance)
R = 15 cm
we know, focal length, f = -R/2
= -15/2
= -7.5 cm
let v is the image distance.
a) use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-7.5) - 1/5.4
v = -3.14 cm
b) magnification, m = -v/u
= -(-3.14)/5.4
= 0.581
image height, h' = m*h
= 0.581*2.9
= 1.68 cm
c) upright
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.