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:
[tex]\frac{1}{q}=\frac{1}{f}-\frac{1}{p}[/tex]
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
[tex]\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}[/tex]
[tex]q=\frac{1}{0.047 cm^{-1}}=+21.3 cm[/tex]
The sign of q is positive, so the image is real.
Now let's 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
By substituting p = 50 cm and q = 21.3 cm, we find
[tex]h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o[/tex]
So we notice that:
[tex]|h_i| < |h_o|[/tex] : this means that the image is smaller than the object
[tex]h_i < 0[/tex] : this means that the image is inverted
so, the correct option is:
It is real, inverted, and smaller than the object.
The image formed by the lens is real, inverted, and smaller than the object.
The distance of the image formed is deterimed by applying lens fromula as shown below;
1/f = 1/v + 1/u
1/v = 1/f - 1/u
1/v = 1/15 - 1/50
1/v = 0.04667
v = 21.43 cm
The size of the image formed is calculated as follows;
M = v/u
M = 21.43/50
M = 0.43
Thus, the image formed by the lens is real, inverted, and smaller than the object.
Learn more about converging lens here: https://brainly.com/question/22394546
