An object placed 30" " cm from a concave lens forms a virtual erect image at a distance of 10" " cm from the lens. The magnification produced by the lens is:

Correct option is D

The correct option is D

Given:

v = image distance = -10 cm

u = object distance = -30 cm (negative sign for object on the left side of the lens)

Formula Used:

​The formula for magnification (m) is:

​$$m = \frac{v}{u}$$​​

Solution:

​Substituting the values:

​$$m = \frac{-10}{-30} = \frac{1}{3}$$​​

Therefore, the magnification produced by the lens is $$\frac{1}{3}$$​.​