Why do things that are far away seem smaller?

Imagine there are many people in a row (all are same height)

    A            Other Guy B             C             D
|
you |  distance     |        distance        distance
| <-------->    |        <------->   |  <------->
|               |                    |             |
|               |                    |             |
------------------------------------------------------  floor


From your point of View you see D much smaller than you, then.. Remember, if this is an "effect" then D is not smaller than you, it's just far away (and this is not earth curvature, because D is smaller in all sense not only in height)

so..

Imagine you have a laser pointer and want to point your partner's face to annoy him so how do you target it? do you align the laser with the floor at the height of your own face believing in geometry? or do you point directly at the "little person" D ?

    A               B                    C             D
|. . . . Laser path . . . . .  . >. . . . . . >. . . . . . >. . .  . >
|  ^            |
you |  |parallel    |                    |
|  |            |                    |             |
|  v            |                    |             |
------------------------------------------------------ floor


Then you don't hit him, and the "visual effect" became so real that you start to think D is for Dwarf.

There is something Wrong? Physics opinions are welcome!

I think the answer is that floor itself will seem not aligned eigther so targeting the Dwarf you will be really aligned with floor because floor itself will have perspective effect.

    A               B                    C             D
|. . . . Laser path . . .
|               |        ''''''''''''......
you |               |                    |      ......>|
|               |                    |      _______|___floor
|               |         ___________|------
--------------------------


Anyway its weird

• Regarding the question: 'do you align the laser with the floor at the height of your own face believing in geometry? or do you point directly at the "little person" D ?' -- If D is the same height as you, then there is no difference between leveling the laser pointer at the height of your own face and pointing directly at 'the little person'. It is the same thing. Jan 21 '11 at 19:48
• It seems a question of projective geometry. Jan 24 '11 at 12:10

I will kindly assume that your question isn't a hoax.

Objects that are further from the observing eyes look smaller because of geometric optics. If an object at distance $D$ from our eyes has size $S$, the rays from its endpoints will arrive to our eyes from the same angles as the rays from a smaller object of size $S/k$ whose distance is $D/k$, because of two simple similar triangles. The angle is essentially $S/D=(S/k)/(D/k)$.

A single eye may only detect the direction from which a light ray is coming - the light ray will create a point on the retina that only depends on the angle from which it is coming (let's neglect focusing now - focusing either by lens in each eye or by the relative position of both eyes is able to determine the distance of close enough objects), so it will think that the large object of size $S$ is as small as the smaller object of size $S/k$ just because the larger object is $k$ times further.

A random picture sufficient to explain what I mean, see below:

In the picture above, "A" is the eye, the line interval "BC" is the closer and smaller object, and the "DE" interval is the larger but more distant object. The light rays from points "C" and "E" arrive to the eye "A" from the same direction, and similarly for the light rays from "B" and "D", so the eye can't really distinguish the objects "BC" and "DE".

Cheers LM

• Hello @Luboš Motl thanks for answering, of course it's not an hoax, but a funny question, you have ignored the target shooting part, anyway you give a good answer +1
– Hernan Eche
Jan 21 '11 at 14:13
• @Luboš Motl: I've added your picture in-text. I hope you don't mind the edit. Jan 21 '11 at 16:23
• Thanks, Frédéric, for your image fix: a great function I should have used - and I may use in the future. Hernan, good that you're still interested. Well, the eye is smaller than most objects we watch but it is not a point. On the contrary, for us to be able to distinguish different directions of the light rays, it's important for the retina to be big enough - to have a nonzero size. The only "point" in the eye is the pupil through which the light is coming. The body wants it to be small for images of objects at any distance to be sharp, but large to get enough light when it's dark. Jan 21 '11 at 18:01
• Of course I am still interested =), I still feel don't have the answer I search, but now its a bit clearer for me, although I don't agree the triangle and straight light is all the answer, I really keep thinking the size of detector matter, let's imagine a really big detector, then I think you not only loose sense of perspective, you could even look "through" objects, If you want to see what's behind your monitor, (if there is enough light) perhaps is enough to stand up of your chair.
– Hernan Eche
Jan 21 '11 at 18:16
• Other thing in favour of the idea of the size of detector is that definition of distance is itself related with the size of the plane of measurement, in fact the distance to an infinite plane can't be defined easily because the plane seem the same at any distance
– Hernan Eche
Jan 21 '11 at 18:51

This is a good question, and what you are encountering is what is often called the inverse r-squared scaling law, or more simply, the inverse square law. In this case, we can write a crude relationship:

$$Size_{apparent} \propto Size_{actual}\dfrac{1}{R^2}$$

If we think in terms of steradians, we can think in terms of the area of spheres we can take the actual two dimensional size of an object (say 1.5m high and 0.5m wide = 0.75 $m^2$); and if we say that object is 3m away, we would compare that area to the area of a sphere that has a radius of 3m.

$$4\pi{R^2} = 4\pi{3^2} = 36\pi$$

$$\dfrac{0.75}{36\pi} = \dfrac{1}{48\pi}steradians$$

If we take that same object and move it out to 30m we find that its apparent size shrinks much faster than its linear distance:

$$4\pi{R^2} = 4\pi{30^2} = 3600\pi$$

$$\dfrac{0.75}{3600\pi} = \dfrac{1}{4800\pi}steradians$$

This is the reason that an object appears very small as you move it further away from you.

If you exclude curvature effects of the earth, the object that is further away from you because its apparent cross section is much smaller than if it were closer.

In answer to your question, barring other physical effects (particularly curvature of the earth effects), if you wanted to hit an object of equal height that was further away with a laser, you would want to aim as straight and parrallel to the ground as possible, and above the ground at an appropriate height.