Answer:
It is real, inverted, and smaller than the object.
Explanation:
First of all, we can use the lens equation to find the location of the image:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}[/tex]
where
q is the distance of the image from the lens
f = 15 cm is the focal length (positive for a converging lens)
p = 50 cm is the distance of the object from the lens
Solving the equation for q,
[tex]\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}\\q=\frac{1}{0.047 cm^{-1}}=21.3 cm[/tex]
The distance of the image from the lens is positive, so we can already conclude that the image is real.
Now we can also write 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.
Substituting p = 50 cm and q = 21.3 cm, we have
[tex]{h_i}=-h_o \frac{21.3 cm}{50 cm}=-0.43 h_o[/tex]
So from this relationship we observe that:
[tex]|h_i| < |h_o|[/tex] --> this means that the image is smaller than the object, and
[tex]h_i < 0[/tex] --> this means that the image is inverted
so, the correct answer is
It is real, inverted, and smaller than the object.
