# Rank the images on their bases of size largest to smallest (physics).?

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The same object is placed at different distances d in front of six different concave spherical mirrors. Each mirror has the focal length f listed below.

1) d=15cm

``  f=20cm``

2) d=10cm

``  f=5cm``

3) d=20cm

``  f=10cm``

4) d=10cm

``  f=20cm``

5) d=15cm

``   f=5cm``

6) d=5cm

``   f=20cm``

I know that as the object moves away from the focal point, in either direction, the image size decreases. However I cant seem to get the right answer I really need some help here.

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• Have two equations:

(1/f) = (1/do)+(1/di)

where f is the focal length, do is the object distance, di is the image distance and

hi = -hodi/do

where hi is the image height

and ho is the object height

1. (1/20)=(1/15)+(1/di)

(1/di) = -1/60

di=-60

hi = 60ho/15=4ho

1. (1/5)=(1/10)+(1/di)

1/di = 1/10

di=10

hi= -10ho/10 = 10 = -ho

1. (1/10)=(1/20)+(1/di)

1/di = 1/20

di=20

hi=-20ho/20 = -ho

1. (1/20)=(1/10)+(1/di)

1/di = -1/20

di=-20

hi=20ho/10 = 2ho

1. (1/5)=(1/15)+(1/di)

1/di = 2/15

di = 15/2

hi = -15ho/2*15 = -ho/2

1. (1/20) = (1/5) + (1/di)

1/di = -3/20

di = -20/3

hi = 20ho/3*5 = 4ho/3

So in order of size we have: 1,4,6,(2,3=),5