Published: april 11th 2011 (monday), 14:32

Yesterday my friend Ćukasz came to me and asked for help with solving some physics problem in optics (you know, lenses and stuff… ;-J ). Here's that problem in a form he gave it to me:

We have an object, a lens, and the real image of that object. The distance between the object and its image is 1 meter, and the magnification is 5 times. What is the focal distance of this lens?

My friend didn't know how to solve this problem; he didn't know either any formulas from optics which could help him to solve it. However, when I draw a picture illustrating this problem, i concluded that **no fancy formulas are needed!** :-) We also won't have to know any physical properties of the lens, such as refractive index between glass and air etc. We don't even need to know the law of light refraction nor any formulas related to it! :-D The only thing we need is **a little bit of geometry** :-)

So let's look how does it appear on the picture (see on the side):

For my own convenience I've placed a grid on my drawing to faciliate noticing the geometrical proportions (actually I draw in a grid-paper notebook).

Firs I drew a horizontal line – the optical axis, and on it I drew a lens, aligned to the grid. Next I drew the object represented by a unit vector (an arrow) at the distance of one square to the left from the center of the lens. Why a unit arrow at a unit distance? To get simpler and more obvious proportions :-)

From the tip of the arrow (object) I drew two straight lines (rays of light) to the lens: one horizontal, parallel to the optical axis, and the secon one straight through the center of the lens and further.

Now it's time to draw the image, which has to be 5 times bigger from the original object. In my picture it has exactly 5 units of length. Its tip has to touch the line which came from the tip of the original object and went through the centre of the lens, because it represents the light sent from that tip, and on the image it also has to aim at the tip. I've placed the arrow of the image in such a place, that it has height of 5 units and touch that line.

Now look carefuly at that line which goes through the center of the lens. Thanks to it we can know where the centre of the lens divides the distance between the object and its image; that is, what is the ratio of these distances: from the object to the lens, and from the lens to the image.

To figure it out, it's enough to notice that we have two similar triangles there! :-) One of them is spanned between the tip of the object, its tail, and the lens's centre. The second one is spanned between the tip of the image, its tail, and the lens's centre. We know that both triangles are similar, because they have exactly the same angles, but one of them is a bigger/scaled version of the first.

How much bigger?

It's enough to look at their sides laying along the arrows :-) The arrow of the image is 5 times bigger than the original arrow, thus we know that the whole triangle is also 5 times bigger. This means that the other sides are also in the 1:5 ratio. So the distance from the centre of the lens to the image is 5 times bigger than the distance from the original arrow to the lens's centre. You can see that on the picture if you count the grid squares: one square between the object and the lens, five squares between the lens and the image. 6 squares in total.

We know from the problem's description that the distance between the object and its image is 1 meter. This means that if we divide this meter with squares, such as on the picture, into 6 equal parts, each part will have 1/6 meter. The lens will be at the distance 1/6 meter from the original object and 5/6 meter from the image. This way we already know the distance of the lens from each of them, in meters. We now only need to calculate the distance of its focus (the red dot) from the lens, that is, its focal distance.

The second line (ray of light), which has came from the tip of the object and went horizontally, and then it has gone into the lens, also has to go into the same place of the image (that is, the tip). Thanks to that we don't need to know the refractive index – we simply are sure, where that light ultimately **must** come ;-J

It's precisely that line which goes through the lens's focus – a place where this line crosses the optical axis. How can we find that place?

Recall that we have cast the image of the object straight onto the lens, parallelly to the optical axis, such that in the place of the lens it will have the same height: the unit height in our case. In that place the ray of light refracts and goes through the focus and further, to meet the former ray of light (that one which has gone thgough the lens's centre) in the place where is the tip of the image's arrow.

We can notice that again we've got similar triangles! This time the first of them is spanned between the tip and tail of the image at the lens and the focus, and the other is spanned between the focus and the tip and tail of the image behind the lens. We can also see that (such as before) one of the sides (the one laying along the arrow of the image behind the lens) is 5 times longer than the other (the unit-length one, laying along the lens), so again both triangles are in 1:5 ratio. Thus the focus also has to divide the optical axis between the lens and the image behind it in the 1:5 ratio, that is, in such a way that the distance of the image from the lens were 5 times bigger than the distance between the focus and the lens.

We already know that the whole distance between the lens and the image is 5/6 meter. We must then divide it further into 6 another parts, to put the focus at the distance of one such part from the lens, and the image five such parts further. Then we'll get the required 1:5 ratio. So let's divide 5/6 meter further into 6 another parts: 5/6/6 (5 divided by 6 and one more time divided by 6) gives us 5/36 meter, and that is the length of one such part. The focus is at the distance of one such part from the lens, so we have the final answer! Here it is:

The focus is at the distance of 5/36 meter from the lens.

Piece a cake, isn't it?! :-D And we haven't need any formulas nor refractive indices for that! Only pure geometrical knowledge :-)

That's how geometry can help us in our lives! Remember about that next time you'll be building a telescope, a camera, or a microscope ;-)

We'll get back to the subject of the geometry of lenses and other optical devices sooner or later. You'll see then that lenses are not an exception here. When you look closer at the formulas from optics you may find in textbooks, then you should notice that these formulas are exactly the same for lenses as for flat and convex/concave mirrors! It can't be a coincidence! :-> You'll see that it results from the fact that any optical devices are governed by exactly the same geometrical laws. You'll also see that this geometry is simply one of the Pythagorean means: the harmonic mean. But I'll tell you about all that next time…

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