What is the magnification of the image?

Question

The focal length of a diverging lens is negative. If f = −16 cm for a particular diverging lens, where will the image be formed of an object located 41 cm to the left of the lens on the optical axis?

What is the magnification of the image?

Answer 1

1/F = 1/d + 1/f; 1/f = 1/F - 1/d; f = dF/(d - F) = (- 16cm· 41cm)/(41cm + 16cm) = - 11.5 cm.

So image will be imaginary, distance from lens by 11.5 sm.

Magnification of image is -11.5/(-16) = 23/32.

In my notaition F = - 16 cm, d = 41 cm