Here's the problem I'd like to solve.
If I'm 1 ft away from a computer screen and a word on the screen appears a certain size, is there an equation or calculation that will tell me how big that word will appear (relative to how large it appears at 1 ft) when I'm 2 ft away from the computer screen? 4 ft away from the screen? 8 ft away from the screen? And so on...
Thanks in advance for your help.
John
On
If we assume that your eye is directly opposite the word on the screen, we get this diagram, where $a$ is the size of the word (relative to 1 foot) and $x$ is your distance from the screen.
We have two right triangles, one with angle $\alpha$ and sides 1 and $a$, and another with angle $\beta$ and sides $x$ and $a$. Using the tangent function we get
$$\frac{\beta }{\alpha } = \frac{{{{\tan }^{ - 1}}\frac{a}{x}}}{{{{\tan }^{ - 1}}a}}$$
as the relative size of the word. Not that I define "size" as the angle made by the object at your eye.
Let us imagine to look at a vertical screen with letters of height $k$ from a $1 $ ft distance. Assume that our eyes are directed horizontally towards the middle level of the letters. Now let us consider the triangle obtained by connecting our eyes to the upper and lower margin of the letters, and whose base is formed by the letters themselves. This is an isosceles triangle with base equal to $k$ and height equal to $1$ ft. The letters are then projected on our retina as an upside-down image. To figure this, we can prolong the two equal sides of the triangle behind the level of our eyes, to form a smaller, similar triangle whose base is given by our retina. If we call $r$ the small distance between the eye lens and the retina, the two triangles are similar with ratio $1/r$.
Now let us move the screen at a distance of $2$ ft. If we prolong the two equal sides of the larger triangle towards the screen without changing the angle between them, we get a new isosceles triangle whose base is $2k$ and whose height is $2$ ft. Since the real size of the letters is still $k$, they will cover only 50% of the new base. The same thing occurs in the smaller triangle inside our eyes, so that the letters on our retina will appear 50% smaller as compared to the perception obtained at the initial distance of $1$ ft. Note that the angle obtained by drawing two new lines from our eyes to the upper and lower margins of the letters at a distance of $2$ ft is not exactly half of the initial angle: its value changed from $\displaystyle\arctan(k)$ to $\displaystyle\arctan(k/2)$. The same occurs for the symmetric angle at the vertex of the smaller triangle within our eyes. However, our perception is not proportional to the changes in the angle, but to those in the projection amplitude on the retina, which has been reduced by 50%.
Generalizing, the "appearance" of letters of real size $k$ observed from a distance $j$, when compared to that obtained at a unitary distance, is given by an apparent size of $k/j$.