Answer:
The image is virtual and upright
Explanation:
The magnification of a lens can be written as follows:
[tex]M=\frac{y'}{y}=-\frac{q}{p}[/tex]
where
y' is the size of the image
y is the size of the object
q is the location of the image with respect to the lens
p is the location of the object with respect to the lens
In this situation, the magnification is positive. This means that:
- y' (the image) has same sign as y (the object) --> the image is upright (same orientation as the object)
- q has opposite sign to p --> this means that the image is located on the same side as the object, so it is a virtual image
