Projectors and Magnifiers

Move an object toward a lens's focal point and its image runs away: farther and farther, without bound. Just outside the focal point the image is real and far away, which is how a projector works. Just inside, it's virtual, which is how a magnifying glass works.

How it works

Infinite limits and vertical asymptotes

In one sentence

$\lim_{x\to a^+}f(x) = +\infty$ means $f(x)$ becomes arbitrarily large for $x$ close enough to $a$ on the right; if either one-sided limit at $a$ is infinite, $x = a$ is a vertical asymptote.

General formula

{\htmlData{sym=1}{d_i}} = \frac{{\htmlData{sym=2}{f}}\,{\htmlData{sym=0}{d_o}}}{{\htmlData{sym=0}{d_o}} - {\htmlData{sym=2}{f}}}, \qquad \lim_{{\htmlData{sym=0}{d_o}}\to {\htmlData{sym=2}{f}}^+} {\htmlData{sym=1}{d_i}} = +\infty, \qquad \lim_{{\htmlData{sym=0}{d_o}}\to {\htmlData{sym=2}{f}}^-} {\htmlData{sym=1}{d_i}} = -\infty

Dividing by a number near zero gives a huge result, and its sign depends on which side of zero you're on.

What the symbols mean

$d_o$
Object distance
How far the object is from the lens.
$d_i$
Image distance
Where the image forms; negative means a virtual image on the object's side.
$f$
Focal length
Where parallel rays meet; the vertical asymptote.

The thin-lens equation $\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$ gives the image distance $d_i = \frac{f\,d_o}{d_o - f}$ .
As $d_o \to f$ , the denominator goes to 0 while the numerator stays near $f^2$ , so $|d_i|$ grows without bound.
From the right ( $d_o > f$ ) the image is real and far away: $d_i \to +\infty$ . From the left it's virtual: $d_i \to -\infty$ .
The line $d_o = f$ is a vertical asymptote. Far away, $d_i \to f$ : a horizontal asymptote.

Learn the math

Infinite LimitsCalc I · Lesson 9, Limits Limits at InfinityCalc I · Lesson 10, Limits