Try out the formula above. Read the threads. Build some lights and measure them. It will all work out. What you are stating is a common misconception, like I said.

Why do you mention a toothpick sized reflector? It obviously needs to be wider than the die of the LED as you also mention.

## Example:

Lets say you have two identical LEDs with a luminance of 200cd/mm^{2}. Each has a die with an area of 1mm^{2}.

Lets also say that you have two parabolic reflectors. Both have an LED hole with a diameter of 10mm. Both have a center depth of 20mm. Both have an aluminium coating with 90% reflectance of visible light. Both have an ar-coated glass lens with 96% transmittance (typical chinese flashlight quality) on top.

Reflector A has an outer diameter of 20mm.

Reflector B has an outer diameter of 50mm.

Now you want to find out how far you can see at night with these two lights (the ANSI range of a flashlight). To calculate this distance you need the luminous intensity [candela]. To calculate luminous intensity you need to calculate the frontal surface area of both reflectors.

### Reflector A:

refl_a_area = area_circle - area_led_hole

= (20mm / 2)^{2} * pi - (10mm / 2)^{2} x pi

= 314.2mm^{2} - 78.5mm^{2}

= 235.7mm^{2}

refl_a_lum_intensity = luminance * refl_a_area * transmission_losses * reflectivity

= 200cd/mm^{2} * 235.7mm^{2} * 0.96 * 0.9

= 40,729cd

Note here that cd is the same as lux@1m.

refl_a_ansi_range = sqrt ( refl_a_lum_intensity / 0.25 Lux )

= sqrt (40,729lux / 0.25lux)

= 403.6m

### Reflector B:

refl_a_area = area_circle - area_led_hole

= (50mm / 2)^{2} * pi - (10mm / 2)^{2} * pi

= 1963.5mm^{2} - 78.5mm^{2}

= 1885mm^{2}

refl_b_lum_intensity = luminance * refl_a_area * transmission_losses * reflectivity

= 200cd/mm^{2} * 1885mm^{2} * 0.96 * 0.9

= 325,728cd

refl_b_ansi_range = sqrt ( refl_b_lum_intensity / 0.25 Lux )

= sqrt (325,728lux / 0.25lux)

= 1141.5m

^{ }

As you can see, the depth of the reflector has no effect on the results.