# Light Bulb Question! **URL:** \ **Category:** Factual Questions\ **Created:** [May 14, 2002, 2:14am UTC](https://boards.straightdope.com/t/light-bulb-question/108945 "2002-05-14T02:14:54Z")\ **Posts on this page:** 14\ **Page:** 1
**Author:** ![rayray5884](https://avatars.discourse-cdn.com/v4/letter/r/57b2e6/32.png) [@rayray5884](https://boards.straightdope.com/u/rayray5884)\ **Post date:** [May 14, 2002, 2:14am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/1 "2002-05-14T02:14:54Z")
My dad and I were installing a light earlier and we got into a arguement about lumens and how much brighter (2) 60 Watt bulbs would be compared to (1) 60 Watt bulb. Question then: If (1) 60 Watt bulb puts off 870 Lumens (as marked on the package), do (2) 60 Watt bulbs both putting off 870 Lumens each put off a combined 1740 Lumens?? or is it brighter than one but not exactly double?? Thanks in advance! ---
**Author:** ![bbeaty](https://avatars.discourse-cdn.com/v4/letter/b/c0e974/32.png) [@bbeaty](https://boards.straightdope.com/u/bbeaty)\ **Post date:** [May 14, 2002, 2:21am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/2 "2002-05-14T02:21:25Z")
The real question to ask: if one light bulb is putting out 870 lumens, and you add a second bulb, how can the first bulb SOMEHOW KNOW that the other bulb is there? In order to reduce it’s output below 870, it would have to know that the other bulb has been added. And if neither bulb knows that the other one is there, then each constantly contributes its 870 lumens separately, and if you want to know the total, you just add them. ---
**Author:** ![CzechHistory411](https://avatars.discourse-cdn.com/v4/letter/c/94ad74/32.png) [@CzechHistory411](https://boards.straightdope.com/u/CzechHistory411)\ **Post date:** [May 14, 2002, 2:35am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/3 "2002-05-14T02:35:18Z")
I _believe_ (and someone who is either smarter than I or else not too lazy to get out of bed and look it up will correct me if I’m wrong) that the light output that reaches a given point varies with the square of the energy of the source. That means that in order for a light to appear twice as bright, you would need four times the wattage; to appear three times as bright, you need nine times the wattage. Two 60-watt bulbs would therefore not appear twice as bright as a single 60 watt bulb, though I’m not sure if you are measuring total light output (which should be twice as much) or the output at a point (which would be less than twice as much). It has to do with the light being spread out over the surface of a sphere. Again, IANALuminologist, and YMMV, especially in parallel universes. – CH ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [May 14, 2002, 5:51am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/4 "2002-05-14T05:51:58Z")
Lumen is a linear scale so yes, if one bulb outputs 870 lumens, two such bulbs would output 1740 lumens. If one light blocks the light of the other then it will be less, but if you can see both bulbs then the lumens are additive. Whether it _appears_ twice as bright is a different question. Human eyes are not that sensitive to absolute brightness. ---
**Author:** ![dtilque](https://avatars.discourse-cdn.com/v4/letter/d/d6d6ee/32.png) [@dtilque](https://boards.straightdope.com/u/dtilque)\ **Post date:** [May 14, 2002, 7:42am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/5 "2002-05-14T07:42:02Z")
Silly me. I thought this was going to be a thread on how many Straight Dopers does it take to change a light bulb. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [May 14, 2002, 11:43am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/6 "2002-05-14T11:43:20Z")
Radiometry is a very weird and annoying branch of optics. The concepts are non-obvious, and don’t even get me started on the units – radiometry and photometry never heard of the metric system, being filled with mutant units like lumens, candelas, nits, stilbs, and phots. Don’t say the units too fast, or you’ll end up with a dirty mouth. A **lumen** is a unit of luminous flux emitted by a uniform, point source of intensity one candela into a cone of solid angle of one steradian. It’s luminous flux, because you “weight” the radiant flux (outflow of photons) by the eye’s response to the photons (so you don’t count infrared photons, of which there are a whole lotta in a light bulb, and you weight red ones less than green ones, since you’re more sensitive to green.). As **scr4** and others have noted, the flux is additive, as long as you can see both filaments. But when you get into other radiometric and photometric quantities, the results aren’t additive. In radiometry and photometryt, for instance, _Intensity_ is defined as flux per unit area of emitter per unit solid angle. ---
**Author:** ![bbeaty](https://avatars.discourse-cdn.com/v4/letter/b/c0e974/32.png) [@bbeaty](https://boards.straightdope.com/u/bbeaty)\ **Post date:** [May 14, 2002, 12:54pm UTC](https://boards.straightdope.com/t/light-bulb-question/108945/7 "2002-05-14T12:54:40Z")
> [@](#): > > I believe (and someone who is either smarter than I or else not too lazy to get out of bed and look it up will correct me if I’m wrong) that the light output that reaches a given point varies with the square of the energy of the source. Nope, light “is” energy. Unless you’re talking about phase-locked beams and interference, light beams just sum together. You’re probably thinking about field amplitude. The energy flux is proportional to the square of the e-field intensity (or, if the EM waves are travelling on wires or along waveguides, the energy is proportional to the square of the voltage, which is almost the same thing.) ---
**Author:** ![CzechHistory411](https://avatars.discourse-cdn.com/v4/letter/c/94ad74/32.png) [@CzechHistory411](https://boards.straightdope.com/u/CzechHistory411)\ **Post date:** [May 15, 2002, 11:07am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/8 "2002-05-15T11:07:37Z")
Was I perhaps confusing lumens with footcandles? I have to admit that my light knowledge is based on photography experience rather than physics class experience. – CH ---
**Author:** ![Delta-9](https://avatars.discourse-cdn.com/v4/letter/d/3be4f8/32.png) [@Delta-9](https://boards.straightdope.com/u/Delta-9)\ **Post date:** [May 15, 2002, 1:33pm UTC](https://boards.straightdope.com/t/light-bulb-question/108945/9 "2002-05-15T13:33:48Z")
> [@](#): > > A lumen is a unit of luminous flux emitted by a uniform, point source of intensity one candela into a cone of solid angle of one steradian. ::slaps forhead:: Oh, **now** I get it. Duh. ---
**Author:** ![Cartooniverse](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cartooniverse/32/3084_2.png) [@Cartooniverse](https://boards.straightdope.com/u/Cartooniverse)\ **Post date:** [May 16, 2002, 1:23am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/10 "2002-05-16T01:23:11Z")
> [@](#): > > \*Originally posted by CalMeacham \* > \*\*A **lumen** is a unit of luminous flux emitted by a uniform, point source of intensity one candela into a cone of solid angle of one steradian. \*\* \<-----------fanning self. Oh mah LAWDY I think I love you !! 😃 😃 😃 😃 Ain’t nuttin’ finah than a fellah who knows his lumenocity… Cartooniverse ---
**Author:** ![Cartooniverse](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cartooniverse/32/3084_2.png) [@Cartooniverse](https://boards.straightdope.com/u/Cartooniverse)\ **Post date:** [May 16, 2002, 1:24am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/11 "2002-05-16T01:24:58Z")
…is it bad that I misspelled Lumins?.. ---
**Author:** ![El\_Zagna](https://avatars.discourse-cdn.com/v4/letter/e/d2c977/32.png) [@El\_Zagna](https://boards.straightdope.com/u/El_Zagna)\ **Post date:** [May 16, 2002, 2:03am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/12 "2002-05-16T02:03:40Z")
I think the problem is the inherent subjectivity in determining what “twice as bright” is. Remember the f-stops on your pre-digital camera? Each stop represented twice as much light as the next - or half as much, depending on which way you are going. So a picture that is taken with one f-stop will be “twice as bright” as one taken stopped down, but you wouldn’t think so by looking at them together. One is just slightly brighter than the other. Same thing with hearing. You have to double the decibals before you can detect any increase in volume. So, yeah, adding another 60-watt bulb, makes the room twice as bright and only slightly brighter at the same time. ---
**Author:** ![ScoobyTX](https://avatars.discourse-cdn.com/v4/letter/s/ec9cab/32.png) [@ScoobyTX](https://boards.straightdope.com/u/ScoobyTX)\ **Post date:** [May 16, 2002, 2:30am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/13 "2002-05-16T02:30:23Z")
Another factor might be the wiring configuration. If the two bulbs are wired in series, the voltage drop through each would be half of the voltage drop through a single bulb. This would reduce the power consumed by each bulb and the light output below the ratings. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [May 16, 2002, 4:11am UTC](https://boards.straightdope.com/t/light-bulb-question/108945/14 "2002-05-16T04:11:48Z")
Pardon me if this is a nitpick, but isn’t flux a measure of intensity per area, rather than intensity per solid angle? At least, that’s the way the term is used in astronomy. Intensity per steradian would just be another measurement of intensity. Of course, I’ve never been able to understand why luminosity isn’t just measured in watts, either, so maybe I’m just missing the point somewhere.