# Distance From Eye VS Visible Size

1. Jul 8, 2011

### smengler

Hi, this should be an easy question unless I'm not understanding it correctly.

As an object gets closer to your eye, how does the "visible" size of the object increase. I'm thinking it can be modeled by either an inverse or inverse square function, but I don't know. Am I on the right path here? I looked on Google but didn't find anything to help me.

2. Jul 8, 2011

3. Jul 8, 2011

### Physteo

hi! Actually you can find an inverse quadratic dependence between the "square of the sine of the half the angle (that represent the visible size) and the distance". Figure out the problem as an isosceles triangle characterized by the unique angle "ϑ", the catetes "a"(could be the distance) and "b"(the real size of the object). You'll find with trigonometry that:
$$sin^2{\theta/2} = \frac{(b/2)^2 }{a^2 + (b/2)^2}$$
could think that in this way:
$$f(ϑ) ≈ \frac{1}{(distance^2+something)}$$

Last edited: Jul 8, 2011
4. Jul 15, 2011

### smengler

Thanks, I'm going to have to look at this for a while before I understand what's going on, but now I'm heading in the right direction. I'm just sort of confused on how the angles work. I'm just trying to find what the "visible" size is at a certain distance away if I know the actual size at a different distance. What do angles have to do with the question? I've taken grade 12 functions and grade 11 physics, so I'm not very good with the math. Thanks for your help.

5. Jul 16, 2011

### BruceW

Its a good link from jtbell.
Think about it this way: the way we perceive the size of an object is to move our eyes from one side of the object to the other, then the angle our eyeball has rotated is how big the object appears to be to us.
So the apparent size of an object is simply the angle that it takes up in our vision. Therefore:
$$tan( \frac{ \alpha }{2} ) = \frac{w}{2d}$$
Where $\alpha$ is the angle the object takes up in our vision (i.e. the apparent size). And w is the actual width of the object, and d is the distance from us to the object.
$\alpha$ is defined as being somewhere between zero and 180 degrees. So when the ratio w/d is larger, the apparent size of the object is larger (as we should expect).
To calculate the ratio of the apparent sizes of two objects in view, simply divide the $\alpha$ of one by the $\alpha$ of the other.

6. Jul 16, 2011

### Staff: Mentor

Its not physics per se but geometry. The concept is called "similar triangles".

7. Jul 18, 2011

### RedX

Your eyes are a thin lens, with focal length 1.85 centimeters. So another way to approach this problem is with the thin lens equation, that gives you the magnification of an object as a function of distance.

Essentially, m=xi/xo, but 1/xi+1/xo=1/1.85, so m=1.85/(xo-1.85). But the object distance will be much greater than 1.85 centimters (at 25 centimeters your eye loses focus due to the fixed length to retina - this is the near point), so you can approximate this as

m ~ 1/xo

So the size of an object gets magnified or diminished as the inverse of the distance to the object, xo.

Of course geometric optics is essentially ray tracing, and you don't need any of this, but it's nice to know that optics equations are consistent with intuition.