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April 19, 2014

April 19, 2014

Posted by **Jon** on Tuesday, April 8, 2008 at 12:46pm.

di= -10

magnification = |di/do|

magnification = |-10/5.0|

magnification = 2

image size = (object size)*(magnification)

image size = (5.0)*(2)

image size = 10 cm (all of that is from the equation drwls gave me)

2)A rod of length 5.0 cm lies along the axis of a concave mirror of radius of curvature 20 cm. The end of the rod closer to the mirror is 15 cm from the mirror. Find the length of the image of the rod.

20/2

10 cm

- Physics repost -
**bobpursley**, Tuesday, April 8, 2008 at 1:55pmYes on 1, and a quick ray diagram ought to confirm it.

On number 2, find the di for each end of the rod, then the difference is the length. I didnt compute it, but your work indicates you handled it as an image perpendicular to the principal axis, which it is not. I will be happy to critique your work on this.

- Physics repost -
**Jon**, Tuesday, April 8, 2008 at 2:35pm2)f=10

di=(10)(5)/5-10

di=-10

di=(10)(15)/15-10

di=30

you said the difference is the length, so it would be 30--10 which would be 30+10=40? or 30-10=20? I got confused on that part

- Physics repost -
**bobpursley**, Tuesday, April 8, 2008 at 2:39pmNot checking your math, if it is -10 and 30, the distance between those points is 40cm.

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