A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}[/tex]
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
[tex]\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}[/tex]
[tex]q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm[/tex]
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
[tex]h_i = - h_o \frac{q}{p}[/tex]
where [tex]h_i, h_o[/tex] are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that [tex]h_i >0[/tex], which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
