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

NetherCraft 0

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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Thanks in advance.

1 Answer

  • 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


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