An erect object is placed on the central axis of a thin lens, further from the lens than the magnitude of its focal length. The magnification is +0.4. This means: A) the image is real and erect and the lens is a converging lens
B) the image is real and inverted and the lens is a converging lens
C) the image is virtual and erect, and the lens is a diverging lens
D) the image is virtual and erect, and the lens is a converging lens
E) the image is virtual and inverted and the lens is a diverging lens

the image is virtual and erect and the lens divergent; therefore the correct answer is C

Explanation:

In a thin lens the magnification given by

      m = h '/ h = - q / p

where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.

It indicates that the object is straight and is placed at a distance p> f

analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.

Consequently the lens must be divergent

The magnification value is

          0.4 = h ’/ h

          h ’= 0.4 h

therefore the erect images

therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C

he kicked it 10 feet straight up into the air? i think this works.

yup its correct its: c) b b