(a) -48.0 cm, diverging
We can use the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
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:
[tex]\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{-12.0 cm}=-0.021 cm^{-1}[/tex]
[tex]f=\frac{1}{-0.021 cm^{-1}}=-48.0 cm[/tex]
The lens is diverging, since the focal length is negative.
(b) 6.38 mm, erect
We can use the magnification equation:
[tex]\frac{y'}{y}=-\frac{q}{p}[/tex]
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':
[tex]y'=-y\frac{q}{p}=-(8.50 mm)\frac{-12.0 cm}{16.0 cm}=6.38 mm[/tex]
and the image is erect, since the sign is positive.
(c)
See attached picture.
