Q: The linear magnification of an object placed on the principal axis of a convex lens of focal length 30 cm is found to be +2. In order to obtain a magnification of -2, by how much distance should the object be moved ?

Sol: In the first case, the magnification is positive which implies that image is erect, virtual and on the same

side of the lens as the object. If a is the object distance then u=-a and v=-2a. fro, the lens formula, we have

$\large \frac{1}{v} – \frac{1}{u} = \frac{1}{f} $

or , $\large \frac{1}{-2a} – \frac{1}{-a} = \frac{1}{30} $

a = 15 cm

So the object is at a distance of 15 cm from the lens. In the second case, the magnification is negative, the image is real, inverted and no the other side of the lens as the object. If b is the object distance, then u = -b and v = +2b.

Hence , $\large \frac{1}{2b} – \frac{1}{-b} = \frac{1}{30} $

b = 45 cm

Thus the object has to be moved through a distance of (45 – 15) = 30 cm away from the lens.