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A 0.39 cm high object is placed 8.6 cm in front of a diverging lens whose focal length is -7.5 cm. What is the height of the image?
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Added Sat, 08 Aug '15
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p=object distance
q=image distance
f=focal length( negative because diverging lens)
1/p+1/q=1/f
1/8.6+1/q=1/-7.5

1/q = -1/8.6 - 1/-7.5 = (-7.5-8.6)/64.5

q = -4.006 cm

magnification equation:
M=-q/p
=4.006/8.6
=.4658

height of image = m * height of object

=.4658*.39

=.182cm
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Added Sat, 08 Aug '15
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