In order for a current to be maintained through a conductor a difference in
1 answer:
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(a) -48.0 cm, diverging
We can use the lens equation:
where
f is the focal length
p = 16.0 cm is the object distance
q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)
Solving for f, we find the focal length of the lens:
The lens is diverging, since the focal length is negative.
(b) 6.38 mm, erect
We can use the magnification equation:
where
y' is the size of the image
y = 8.50 mm is the size of the object
Substituting p and q that we used in the previous part of the problem, we find y':
and the image is erect, since the sign is positive.
(c)
See attached picture.
