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aes123_gw

Question about light intensity and distance

aes123
17 years ago

I have a relatively simple question: how does light intensity drop with distance? I'm sure I knew this when I was in college physics, but I haven't had to know it since...

I'm beginning plans for a 24"x18"x24" terrarium, and am trying to figure out the lighting. I'd like the lowest plants to recieve 1000-1500fc, so if the lights are 16-20" away, how much light do I need at the top?

Also, if you have suggestions for a 24"x18" terrarium hood, I'd be much obliged.

Thanks in advance.

Comments (3)

  • shrubs_n_bulbs
    17 years ago

    There is no useful relationship you can apply between this type of light and distance. Too much depends on your reflector setup and the exact arrangement of bulbs or tubes. You should calculate the intensity of light you need, multiply it by the area you wish to light, add maybe 20%-50% depending on how much light will never make it to the plants, and then you know how much you need.

    Terrariums are inefficient at getting all the light to the plants because they have glass sides. Three square feet at 1,500fc, plus 50%, means you need 6,750 lumens.

  • ralleia
    17 years ago

    You should be able to use the inverse square law to determine the illuminance. E (illuminance) is given in footcandles.

    E = I/d^2

    Where I is luminous intensity (in candelas) and d is distance. Footcandles are lumens per sq foot, so d should be in feet.

    Do you have light fixtures rated in fc? Usually the only provide lumen output.

    If you try and can't arrive at a reasonable answer, let me know and I'll try to help you figure it out. It'll be good practice for my lighting final this Friday.

  • shrubs_n_bulbs
    17 years ago

    You should be able to use the inverse square law to determine the illuminance

    You can't. The inverse square law applies only to point source lights. A fluorescent tube is not a point source, plus it is always used within a reflector, hence the inverse square law is doubly useless. Imagine a searchlight, a light source with a very accurate reflector. No way does the inverse square law apply, in fact the illuminance decreases only very slowly with distance.

    The inverse square law is simply a special case of the more general rule that the illuminance is equal to the total emitted light divided by the area that it illuminates. If it radiates equally in all directions, then the illuminated area is the surface of a sphere of radius equal to the distance from the light source. The surface area of the sphere is proportion to the square of the radius, hence the illuminance is inversely proportional to the square of the distance from the light. This is the derivation of the inverse square law. Might it possibly be on the final?

    Applying the more general rule to the real world situation of a fluorescent tube (or multiple tubes or bulbs ) focussed by a reflector or system of baffles to only illuminate a given area, then the illuminance in foot-candles is simply equal to the total lumens divided by the area in square feet. No reflector is perfect and some of the light will always be lost and not arrive in the desired area, as little as 10% in some fittings, but generally more. You don't need to know the geometry of the light system, or make any assumptions about inverse square laws or other laws, you just need to know how much light you have and how much of it gets to the area you are interested in. Lumens divided by square feet gives foot-candles.

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