Answer:
It is real, inverted, and smaller than the object.
Explanation:
Let's start by using the lens equation to find the location of the image:
![\frac{1}{q}=\frac{1}{f}-\frac{1}{p}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bq%7D%3D%5Cfrac%7B1%7D%7Bf%7D-%5Cfrac%7B1%7D%7Bp%7D)
where we have:
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, we find
![\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bq%7D%3D%5Cfrac%7B1%7D%7B15%20cm%7D-%5Cfrac%7B1%7D%7B50%20cm%7D%3D0.047%20cm%5E%7B-1%7D)
![q=\frac{1}{0.047 cm^{-1}}=+21.3 cm](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B1%7D%7B0.047%20cm%5E%7B-1%7D%7D%3D%2B21.3%20cm)
The sign of q is positive, so the image is real.
Now let's also write the magnification equation:
![h_i = - h_o \frac{q}{p}](https://tex.z-dn.net/?f=h_i%20%3D%20-%20h_o%20%5Cfrac%7Bq%7D%7Bp%7D)
where
are the size of the image and of the object
By substituting p = 50 cm and q = 21.3 cm, we find
![h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o](https://tex.z-dn.net/?f=h_i%20%3D%20-%20h_o%20%5Cfrac%7B21.3%20cm%7D%7B50%20cm%7D%3D-0.43%20h_o)
So we notice that:
: this means that the image is smaller than the object
: this means that the image is inverted
so, the correct option is:
It is real, inverted, and smaller than the object.