I have been messing around with magnifying glasses recently and I am stuck on a simple concept. From the thin lens equation we know that $$\frac{1}{s\_0}+\frac{1}{s\_i}=\frac{1}{f},$$ with the sign convention of distances relating virtual images or objects being negative, $0 < f$ for convex lenses and $f < 0$ for concave lenses. From playing around in MS Excel, we see that for a lens with $f=28 \; cm$ the plot $m(s\_0)=\bigg[\frac{s\_i}{s\_0}\bigg](s\_0)$ is obtained as below
[![enter image description here](https://i.stack.imgur.com/cO6Ls.png)](https://i.stack.imgur.com/cO6Ls.png)

Now, this tells us that a virtual image is always erect and a real image is always inverted and that the closer it is from the focal point, the bigger the image. Now my main question is this: When we look from a magnifying glass an object that is located at, say $20\;cm$ from the lens, then the object will appear huge, in fact the virtual image will be formed $70\;cm$ in the object space, that is, $70 \; cm$ away from the lens, but what is this equivalent to? My first hunch was "it is as if, if there were no lens, we would see an object 3.5 times bigger than the original 70cm away from the spot where the lens should be" but that is not right, because if that was the case then my brain would process that object as smaller than the original size times 3.5 due to the distance that it was from me. So, what is it equivalent to? How should we think of virtual images?

Additional question: Why is it that when we look from a magnifying glass, if the image is virtual, we don't need to look for a *focus* distance, that is, if I look from $5 \; cm$ or from $20 \; cm$ the image is still sharp, but if the image is real (object away from focal length) then, I have to search for this sweet spot? (I understand the latter but not the former).