A needle of height 6 cm is placed at a distance of 20 cm in front of a lens with a power of –2.5 D. The height of the image of the needle will be:

Correct option is A

The correct answer is (A) 4 cm

Magnification (M) is the ratio of image height (h') to object height (h), and it's also related to object and image distances:

$$m=\frac{h_i}{h_o}=\frac{-v}{u}$$​

Given,

$$h = 6 cm, {u = -20 cm}, v = \frac{-40}{3} cm$$​

$$\frac{h_i}{6} = \frac{-(\frac{-40}{3})} {-20} = \frac{-2}{3}$$​

$$h_i = {6} \times \frac{-2}{3} = -4 cm$$​

The negative sign indicates that the image is upright (virtual images formed by diverging lenses are always upright).