# How to think about virtual images?

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 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).

• What are you plotting? – Not_Einstein Mar 16 at 22:19
• It's the magnification as a function of the distance of the object to the lens – Bidon Mar 16 at 23:54

The lens is like a window into a different world. In that world, your object is 3.5 times larger than the original real-world object. It is also located 70 cm from the window. If you move your head left and right a bit, you will notice that the parallax effects really are as if the object is farther away.

This linear magnification does not tell you how much larger an object will appear to you, i.e. how much more detail you will be able to see. It tells you how large the image is (the transformed object you see in this other world), wherever it's located.

Edit: I missed the other part of your question

Even with a real image, you don't need to search for a 'sweet spot'. The generated image just needs to be far enough away from you so your eye can actually focus on it.

For instance, if you put the object at 30 cm from the lens, the image will appear at 450 cm from the lens. Let's say your eye can only focus on objects 7 cm away from you. This means you need to stand at least 457 cm away from the lens for your eye to focus on the image.

So you either need to observe from more than 4.57 meters away or put the object a bit further away from the lens so the image distance shortens.

(As a side note, you will not be able to put the image less than 28 cm this way. This is because an object infinitely far away will produce the image at the focal point.)

With virtual images, you don't usually have this problem with focusing because the image is, by definition, on the other side of the lens. You rarely put your eye that close to the lens and the image is even farther away.

But if you put the object at 1 cm from the lens, the virtual image will also be ~1 cm in front of the lens. If you put your eye really close to the lens, you will not be able to focus even on this virtual image.

magnification does not involve the size of a picture except if you project ist on a screen , but then if you look fron a large distance she picture in your eyes get smaller and smaller.How large you see something real or virtual ist the angle under wich you see it, and your brain adjusts for the distance. If you look thru a lens, your ey usually adjust to the distance of the paper you look at, not to the distance of 1m or 70cm where the virtual picture is calculated or constructed, so to know the magnification you have to calculate how much your viewing angle is enlarged.(With a lens with f=28cm you will not get much magnification )

"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."

Your doubt arises because you are considering this in two steps - first moving the (image of the) object 70cm away, at which point it would look smaller, and then magnifying that smaller (image of the) object by a factor of 3.5. Your first hunch is correct. As the image recedes from the lens with smaller and smaller object distances, the magnification increases.

You could turn your reasoning around and say your brain would process the larger image as the object being closer to you, then that larger object moved 70cm away, at which point it would look smaller than 3.5x the object. That would also not be correct.