0
$\begingroup$

I was adjusting the position of our projector on our SLA printer and have found that at further distances our models have less adhesion.

Does the inverse square law of light apply to DLP projectors?

I am aware that the inverse square law only is 100% applicable when you have a point as your light source, but I assume close to the same holds constant for other real world light sources.

Any ideas as to to what extent this is true or not?

$\endgroup$
1
  • $\begingroup$ Consider to spell out acronyms. $\endgroup$ – Qmechanic May 7 '20 at 1:37
2
$\begingroup$

The usual way of picturing the inverse square law is to imagine a sphere of light emanating from a point. Since the same about of energy is spread over an increasing area, and that area increases with the square of the radius, the intensity of light must decrease as the square of the radius. However, this is also true for any portion of the sphere as well.

Portion of sphere for inverse square law

Picture lifted from here.

You can think of a directed light source, like a projector or a laser, as just a portion of a spherical light source, with the rest of the sphere blocked off. Even lasers diverge as the travel long distances, so they are subject to the inverse square law as well. There is actually no such thing as a perfectly collimated light source (though exotic sources like Bessel beams come close).

$\endgroup$
1
  • $\begingroup$ When I was in high school, my physics teacher used this in terms of a "butter squirter." A restaurant had a machine that melted a pat of butter and squirted it onto a slice of toast at a distance r. The restaurant got cheap. They moved back to 2r and used the same pat of butter on 4 pieces of toast. Then really cheap... $\endgroup$ – mmesser314 Jun 30 '17 at 4:11
1
$\begingroup$

If you project an image of a square, does the side of the square grow linearly with distance? If so, the area of the square grows as the square of the distance.

The power the projector uses to light up the interior of the square stays the same, so the power per square foot of image goes as $1/r^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.