Final answer:
The focal length of the lens is approximately -26.5 cm, indicating that it is a diverging lens, as the virtual and upright image suggests for an object is placed 13.9 cm in front of a lens, an upright, virtual image is formed 29.3 cm away from the lens.
Step-by-step explanation:
To find the focal length of the lens that forms an upright, virtual image, we use the thin-lens equation which is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance and di is the image distance.
Since the image is virtual and upright, we know that the lens is a diverging lens and the image distance will be negative.
We can substitute the given values, remembering to convert the image distance to a negative because the image is virtual and on the same side of the lens as the object.
do = 13.9 cm (object distance), di = -29.3 cm (since the image is virtual), substituting in the equation:
1/f = 1/13.9 cm + 1/(-29.3 cm)
1/f = 1/13.9 - 1/29.3
1/f = (29.3 - 13.9) / (13.9 * 29.3)
1/f = (15.4) / (407.17)
f = 407.17 / 15.4
f = -26.5 cm approximately
Since the calculated focal length is negative, this confirms that the lens is indeed a diverging lens.