I have a closed box with a hole in the wall. A IR beam is directed at the front of the box. Due to a small 6mm borehole in the wall only a certain part of the beam can pass trough.

The left small pieces are IR sensors. I'm wondering now if I need an additional tube for focusing the beam like depicted?

Application is a light barrier and therfore the beam needs to be not wide but very narrow. 2nd question is: is there a difference if the ir sensor is placed like the right one or like on the left position? enter image description here

Light source TSAL6100 enter image description here

  • $\begingroup$ What is the divergence of the beam? Judging from your HUGE hole diameter, $d=6mm$, the divergence is negligible. $\endgroup$ – Semoi Jun 20 '20 at 7:01
  • $\begingroup$ To be honest I don't know exactly. I attached specs of the light source. The light beam could be 2m-10m far away from the sensor. When lessen the hole diameterI it get's less sensitiv and therefore I need more light(power) I assume? $\endgroup$ – user3435167 Jun 20 '20 at 9:12

If the light source is more than $D=2m$ away, the hole with diameter $d=6mm$ samples a "small angle" of the $4\pi$ solid angle, and the light is in the regime where a geometric description using "rays of light" is applicable. Thus, you can estimate the diameter of the light ray at the detector by using geometric optics $$ \frac{d/2}{D} = \frac{d/2+x}{D+z} $$ enter image description here Compare the result to the size of your detector.

  • $\begingroup$ Thanks that sounds good! z is in this case my tube, right? Is it also possible to calculate how wide is the part of the beam that needs to be cut by a subject that crosses the light beam so that there is no more light at the sensor? The sensor area is 5mm diameter $\endgroup$ – user3435167 Jun 20 '20 at 11:43
  • $\begingroup$ Please draw a picture and check the rule of proportion by yourself. Don't rely on my calculation. To your second question: Of course you can do that calculation using geometrical optics. Both calculations are very similar. If you have done the first one, you just have to change some numbers. $\endgroup$ – Semoi Jun 20 '20 at 12:59

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