Question

The image of an object is formed $3$ times by a convex lens. On interchanging the positions of object and screen, what will be the magnification in that case?

Answer 1

As we know the fact that if we put the object between F and 2F (where F =Focal Length) the image enlarges and formed beyond 2F point for convex lens.

and Suppose if the object farther than the 2F distance the image diminishes and formed between F and 2F.

So if the image enlarges 3 times, then the object have to be in between 2F and F and image formed beyond 2F .

Let height of object be 'x' therefore image will be '3x' beyond 2F distance.

Second Case

We replaces there positions, that means,

We put the object beyond 2F and hence from our knowledge we know that the image will be formed between F and 2F and diminished.

and Diminished by the same factor by which it is increased in first case.

therefore,$\text{The Magnification is =}\frac{1}{3}$

You can also recheck the answer by putting a $\frac{x}{3}height$ object in the place between F and 2F. It will form a magnified image of $x$ height.