A) -48.0 cm
In order to find the focal length of the lens, we can use the lens equation:
where
f is the focal length
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we have:
p = 16.0 cm
q = -12.0 cm (the negative sign is due to the fact that the image is on the same side as the object, so it is a virtual image, so the sign of q is negative)
Substituting, we find f:
B) Diverging
We have two types of lenses:
- A converging (convex) lens is curved outwards in its center, so that the incoming rays of light parallel to the principal axis are focused into the focus of the lens, on the opposite side
- A diverging (concave) lens is curved inwards in its center, so that the incoming rays of light parallel to the principar axis are deviated away from the principal axis, and they appear to come all from the focal point of the length on the same side of the object
A converging lens is identified by a positive focal length, while the focal length in a diverging lens is negative. Here, f = -48.0 cm, so this is a diverging lens.
C) 6.38 mm
We can answer this part of the problem by using the magnification equation:
where
y' is the size of the image
y is the size of the object
Here we have:
y = 8.50 mm
p = 16.0 cm
q = -12.0 cm
So we find:
D) Erect
We can determine the orientation of the image by looking at the sign of the size of the image found in part C). In fact:
- if the image is erect, the sign of y' is positive
- if the image is inverted, the sign of y' is negative
In this situation, we see that
y' = 6.38 mm
Which is positive, so the image is erect.