Answer:
Distance: -30.0 cm; image is virtual, upright, enlarged
Explanation:
We can find the distance of the image using the lens equation:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]
where:
f = 15.0 cm is the focal length of the lens (positive for a converging lens)
p = 10.0 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving for q,
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm[/tex]
The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).
The magnification can be found as
[tex]M=-\frac{q}{p}=-\frac{-30}{10}=3[/tex]
The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.
Also, the fact that the magnification is positive tells us that the image is upright.
