# Talk:Light meter

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## illumination use

I added illumination use, because light meters are very often used in automatic regulation of lighting.

There is, however, a question of combining different articles related to that: 'PIN diode' or 'light regulation', which is currently, quite unappropriately, redirected to 'Dark-sky movement' article

Mitjaprelovsek 16:56, 29 April 2006 (UTC)

## Exposure determination with a neutral test card

I added the math to this section in attempt to show the basis for readings obtained with a neutral test card. Almost as this was added, however, it struck me as adding more clutter than utility. Unless someone objects, I'm inclined to remove it. Thoughts? JeffConrad 22:41, 15 October 2006 (UTC)

It's part of the article that people will skim-read, so it's not important. But what does "The instructions also recommend that the test card be held vertically and faced in a direction midway between the Sun and the camera" mean? I cannot picture this in my head. -Ashley Pomeroy 22:41, 16 March 2007 (UTC)
Example: imagine an outdoor scene in which the Sun is 30° to the left of the camera. The Kodak instructions would have the test card held vertically, with the perpendicular to its surface 15° to the left of the camera. In my experience, this doesn't always work as expected. The primary problem seems to be that a typical test card is far from a perfect diffuser, so that specular reflections can sometimes have significant effect on meter readings. But the instructions are what they are, so that is how I described them.
No one has objected to the deletion of the math, so I assume that I probably should not have included it in the first place. JeffConrad 00:24, 17 March 2007 (UTC)

## Style consistency

A minor note: the remainder of the article is in American English, so I think "analog" might be preferable to "analogue". There are far greater issues, though; this article has had many contributors, and it reads as a bit of a hodgepodge. It would benefit considerably from an overall edit. JeffConrad 00:24, 17 March 2007 (UTC)

This article could use a lot of work. Much more needed about the history of light meters. How well the various sensors/instruments cover how wide a range of lighting levels. Typical costs. Much more about non-photography uses of light meters, in lighting design and test, scientific uses. Links to such sources.-69.87.204.232 12:37, 19 May 2007 (UTC)

## Lumens

How does one actually measure lumens?-69.87.204.232 12:37, 19 May 2007 (UTC)

Lighting manufacturers use a goniophotometer, which is a mirror on a robotic arm that bounces light into a sensor. Measurements are taken at regular intervals around the lamp or fixture, and the lumen output for a given steridian is then calculated. That's my understanding, anyway. I agree that this could be expanded upon.Ohnoezitasploded (talk) 03:59, 2 October 2010 (UTC)

Here we're really more concerned with illuminance per unit area (lux in the SI) than with the total output (lumens in the SI). The measurement of the output of luminaires (where the total output of interest) is discussed briefly in Photometry_(optics). JeffConrad (talk) 04:34, 2 October 2010 (UTC)

## Light level chart

Lux is the unit of measure most often used to measure light. (The USA still measure in foot-candle*, 2500lux = 26900 foot-candle). Here are some typical lux figures: [1]

• moonlight - <1lux
• living room indoors at night (in front of the TV) - 50lux
• well lit kitchen or office - 500lux
• turn your car lights on in the middle of the day so other people see you coming - 1000lux
• MINIMUM to treat SAD (typically in 2hours) - 2500lux
• clear spring morning, 30 minutes after sunrise in UK (also High Speed 30minute SAD treatment light)- 10,000lux
• bright summers day in UK - 50,000lux
• a high mountain in Kenya (on the equator) at midday - >100,000lux

-69.87.204.232 13:39, 19 May 2007 (UTC)

## Plant growing light meters

Plant growing light meters are readily available, starting at about $20 USD online: Hydrofarm Light Meter "This Hydrofarm Plant Light Intensity Meter is an easy and accurate way to measure natural, fluorescent, or HID light for gardening use. With a range from 0 - 5000 foot-candles you can monitor your lamp's intensity for seasonal light changes in your home. HF's Light Meter is permanently calibrated, accurate to 2%, and requires no batteries. Also includes a booklet with optimum footcandles for a variety of plants." [2] [3] Rapitest Moisture/Light Meter "Manufacturer: Luster Leaf (#1830) UPC: 035307018304 Retail:$19.99
The ultimate tool for measuring two of the most important elements for healthy plants - light and water. Instantly switches from moisture to light meter. No batteries required." [4] -69.87.204.232 13:51, 19 May 2007 (UTC)

## General purpose

Industrial & Commercial Grade Precision Light Meters, including Footcandle Meters, Lux Meters and UV (UltraViolet) Meters $50-150 [5] LX-1010B Economic Lux meter ever made$40 USD Specifications: To check the level of bright Range:0- 2,000/20,000/ 50,000 Lux (+-5%+2D) Accuracy: 5.0% Resolution: 1 Lux Sampling time:0.4 second Dimensions: Body: 4.6 X 2.7 X 1.10" Sensor: 3.26 X 2 X0.8" Weight : 160 g. Measure 0 -- 50,000 lux for a wide range of use. Ideal for use by architectures, light designers, and photographers.. Weight: (Lbs) 0.36LB [6] -69.87.203.79 17:31, 19 May 2007 (UTC)

International Light made scientific radiometers, and they have now merged with Gilway, in Massachusetts US. Specialized units called "belt monitors" are sold for monitoring UV curing facilities. This page has a number of detailed radiometer detector spectral curves, from 200-2000 nm.[7] It would be great to have details about photo detectors of various types in WP.-69.87.203.9 11:55, 30 May 2007 (UTC)

## Consistency of incident-light metering

Is it right to remove the comment about the consistency of incident-light measurements? Although I question whether they are more precise (or more "correct"), I think they usually are more consistent than reflected-light measurements of a given scene. It might be helpful to hear from a cinematographer on this. JeffConrad 21:09, 26 May 2007 (UTC)

## Headroom in digital sensitivity

According to this article, the ISO film speed S is given by S=Kn^2/Lt, where L is the luminance of a medium-brightness area (candela/m^2), n the aperture number, K a constant (usually between 12 and 14), and t is the shutter speed. Usually, medium brightness is taken to be an 18% reflective card. However, this doesn't say anything about the headroom. Should a sensor with a given ISO number be at exactly 18% of saturation for a given luminance L? Or is there a standard amount of headroom? I've searched a bit and I found numbers of 106%, 170%, and 141% for the saturation level. What does the ISO document say about it? This information seems to be missing. See e.g. [8], [9], [10]. Han-Kwang (t) 15:21, 10 October 2007 (UTC)

Solving for S doesn't mean the meter is telling you that that's a definition of S, or a recommendation for S. I mean, yes, given a luminance and a meter calibration, etc., you can back-solve for S. But there's nothing there that says the L value is taken from a "medium-brightness area", is there? It's from wherever you aim the meter. Now, there is actually a saturation-based digital camera ISO speed definition that works something like you say, but it has its own constant, not dependent on any particular meter calibration of course, and leaves about a half-stop of headroom above a diffuse white when average scene is something 18%, if I recall correctly. But nobody rates their cameras by this definition anyway, so don't expect to find any such relationship in practice relative to any ISO setting or rating on a digital camera. If you want to get technical ISO 12232 says "The saturation based speed, Ssat, of an electronic still picture camera is defined as: Ssat = 78 / Hsat, where Hsat is the minimum focal plane exposure, in lux seconds, that produces the maximum valid (not clipped or bloomed) camera output signal." This may not be trivial to interpret as you want, but compare this to the noise-based ISO definitions of the form Snoise40 = 10 / HS/N40 where that H level is in some sense a "mid gray"; the ratio of saturation to this mid-gray is thus 7.8, which is more than 1/0.18 by a factor 1.404, or about a half stop. I hope this helps. Dicklyon 18:53, 10 October 2007 (UTC)

I see, so the equation n^2/t = LS/K is a simply a definition of what the reading that a light meter should produce for a given luminance, even though K is somewhat variable. But as a photographer, of course I would want to know how much headroom I will get if I point the light meter at a 18% card. As you say, officially the saturation level should be 141% (1/2 stop above 100%), but in practice YMMV (e.g. my digital compact camera seems to aim for 18/100 rather than 18/141). Maybe this subject really belongs on film speed, where I also asked this question. Anyway, I think the focal plane exposure is given by H=pi L t/4 n^2 (for objects far away from the lens), so Ssat = 78/Hs = 312 n^2/ (pi*Lsat*t) where Lsat is the luminosity that just saturates the sensor. I guess that 78 comes from 100*pi/4, so Ssat=100 n^2/(Lsat*t), which seems to be an elegant equation -- would it be appropriate to mention on film speed? Han-Kwang (t) 23:43, 10 October 2007 (UTC)

I've posted something on Talk: film speed. Now that I have read Dicklyon's posts here I see that my post there is a re-write of the end of his previous post - ie that the 78 factor in the Ssat formula comes from 141/18. The formula for focal plane exposure used in 12232 (Eq 2) takes flare exposure (Hf), cos^4 loss, lens transmission factor (T) and vignettng factor (v) into account. The pi/4 combines with the other losses (T v cos^4theta, with Hf << H) to produce 65/100 (which is 0.83 pi / 4), so the formula for arithmetic mean focal plane exposure is 65 La t / (100 A^2) where La is the arithmetic mean scene luminance and A is your N. On axis the focal plane exposure would be 0.88 pi L t / (4 A^2). Helen Bach 22:47, 11 October 2007 (UTC)
I propose we continue this discussion on Talk:Film speed. Han-Kwang (t) 11:26, 12 October 2007 (UTC)

## Spectral Sensitivity

I wonder if the paragraph on spectral sensitivity (near the bottom of the section Use in photography) isn't a bit misleading, for two reasons:

1. We imply that most meters are well matched to the spectral sensitivities of most films.
2. We imply that correction of a mismatch is usually straightforward.

What constitutes a “good match” is somewhat subjective; if it means essentially that most meters recommend reasonably acceptable exposures most of the the time, then the match probably is good in most cases. But various meters have wildly differing spectral responses. For example, Pentax spotmeters have very broad response, extending from slightly into UV to well into IR; Kenko (formerly Minolta) have rather narrow response, approximately that of the 1931 CIE photopic observer. In many cases, reasonable results can be obtained with either meter, although relative readings change a bit when measuring objects of different colors. In extreme situations, the differences can be astounding. For example, in a yellow-light room used for semiconductor photolithography, I have seen indications by Pentax and Minolta meters differ by 5 EV; presumably, at least one was incorrect ...

Correcting the response by filtration may work quite well on a meter with broad spectral response, but is less effective on a meter whose built-in filtration already attenuates the response in the region of interest for a particular film.

Much time and effort has been expended in numerous forums arguing about differences in nominal calibration among various manufacturers and about alleged “18% calibration” vs. “12% calibration,” but either difference is often small in comparison with differences in spectral sensitivity. JeffConrad (talk) 19:58, 6 February 2008 (UTC)

## Illuminance, Reflectance & Luminance

In the formulae for Light_meter#Calibrated_reflectance, would it be appropriate to replace ${\displaystyle \zeta }$ with ${\displaystyle \rho }$ for consistency with reflectance or R for consistency with Reflection_coefficient#Optics ? -- Redbobblehat (talk) 17:51, 5 February 2009 (UTC)

I don't think ρ would be a good choice, because it's commonly used for luminance coefficient, where
${\displaystyle \zeta =\pi \rho }$.
I took ς from ANSI PH3.40-1977; it doesn't seem to be used in ISO 5763, which superseded PH3.40, so I guess there's no reason we can't use R (for consistency with Ray). JeffConrad (talk) 21:33, 5 February 2009 (UTC)
Good. R works for me :) --Redbobblehat (talk) 16:50, 6 February 2009 (UTC)
You may note that Ray (2000, 319) gives
${\displaystyle R={\frac {K}{C}}}$
without the factor π because he uses a comparative unit (footlambert) for luminance. JeffConrad (talk) 21:33, 5 February 2009 (UTC)
Hmm yes ... and on the same page, the bizarre EV diagram uses "Scene Luminance (ASB)" - it took me a while to realise these are apostilbs and not alien space bats (much to my disappointment ;) But I take your point that the π factor representing cosine reflectance is "built in" to the some (non-SI) luminance-illuminance-paired units. Perhaps these units have fallen out of favour because they assume cosine reflectance geometry rather than coloidal...? --Redbobblehat (talk) 16:50, 6 February 2009 (UTC)
The presence or absence of the factor π has nothing to do with Lambertian reflectance—it simply indicates the choice of direct or comparative units of luminance. I agree that the EV diagram is nearly incomprehensible. JeffConrad (talk) 22:56, 6 February 2009 (UTC)
I think I may have figured out why 1 lux incident is diffuse-reflected as 1/π nit rather than 1/2π or 2π ...
1. Assume a perfect diffuse reflector disk of area 1m2 which forms the base of a perfect transparent hemispherical dome. The surface area of the dome is twice that of the disk (ie. they have the same radius and origin; circle = πr2 and hemisphere = 2πr2).
2. When 1 lumen/m2 of illuminance is diffuse-reflected off the 1 m2 disk, scattering spreads the lumen evenly (isotropically?) over the inside of the dome. Because the surface area of the dome is 2x that of the disk, the Luminous flux density[11] here is halved to 1/2 lumen/m2.
3. By definition, 1 candela/m2 == 1 lumen/steradian/m2 and a hemisphere subtends a solid angle of 2π steradians. So the luminance emitted through the (transparent) dome is (1/2) lumen/m2 * (2π) steradians = π lm/sr/m2 == π cd/m2 == π nits.
Comments very much welcome. --Redbobblehat (talk) 02:17, 8 February 2009 (UTC)
It's not quite that simple—the flux isn't evenly distributed; the luminance varies as L(θ) = L cos θ, and the infinitesimal flux dM = L(θ) dω, where dω is the infinitesimal solid angle, must be integrated over the hemisphere. On a sphere of radius r; the solid angle dω is subtended by an area dA. A common representation of dA is a ring of radius r sin θ and height r dθ; the area of the ring is r2 sin θ dθ. Then
${\displaystyle d\omega ={\frac {\mathrm {d} A}{r^{2}}}={\frac {2\pi r^{2}\sin \theta \,\mathrm {d} \theta }{r^{2}}}=2\pi \sin \theta \,\mathrm {d} \theta }$
and
${\displaystyle M=\int _{0}^{\pi /2}L(\theta )\,\mathrm {d} \omega =2\pi L\int _{0}^{\pi /2}\sin \theta \cos \theta \,\mathrm {d} \theta =2\pi L\left|{\frac {sin^{2}\theta }{2}}\right|_{0}^{\pi /2}=\pi L\,.}$
JeffConrad (talk) 09:51, 8 February 2009 (UTC)
Now I wish I understood calculus! so thank you for keeping the 2πL separate until the last stage :) Unless I'm missing the point (which is very likely) the formula seems to indicate that, within the specified parameters, the average value of cos(θ) is 0.5 ? Surely cos(θ) comes from lambert's cosine law ? Could this be what Delbruck means by "Note the cos(θ) goes away if you use projected surface area, luminance is then lumen per steradian per projected m2."[12]? ... So if we remove the word "evenly" from stage 2 of my description above, it would all be correct ? --Redbobblehat (talk) 22:39, 9 February 2009 (UTC)
The cos θ indeed comes from Lambert's Law, but the average value of cos θ between 0 and π/2 is 2/π; it's not as simple as multiplying an average value by 2π. JeffConrad (talk) 05:18, 10 February 2009 (UTC)
A footlambert of luminance is contrived to correspond to 1/π footcandles of illuminance so that The luminance of a perfect Lambertian diffuse reflecting surface in foot-lamberts is equal to the incident illuminance in foot-candles. (footlambert), but I'm struggling to see see why an apostilb of luminance = 1/π nits of luminance would be useful ... something to do with comparing specular and diffuse reflections ? -- Redbobblehat (talk) 16:50, 6 February 2009 (UTC)
The apostilb was the mks comparative unit of luminance; i.e., a perfect Lambertian diffuser with illuminance and luminous exitance of 1 lm/m2 has luminance of 1 asb. JeffConrad (talk) 22:56, 6 February 2009 (UTC)
Speculations aside, I propose we (either/or) :
1 Use SI units consistently in the article; relegating the bats, etc to esoteric footnotes, which can also emphasise the necessity of pairing luminance to illuminance units correctly - ie lm/m2 with cd/m2, fc with fL, ?! with apostilb, etc ? (Illuminance[13] and Luminance[14] unit converters may also be useful). --Redbobblehat (talk) 16:50, 6 February 2009 (UTC)
2 Give a whole paragraph to 'how, when and why to apply 1/π to represent lambertian reflectance' --Redbobblehat (talk) 16:50, 6 February 2009 (UTC)
As nearly as I can tell, we have used (or implied) SI units. In any event, the proper place to cover photometric units is photometry (optics) rather than here. JeffConrad (talk) 22:56, 6 February 2009 (UTC)

## Integrating sphere

The mention of an integrating sphere isn't really correct, for two reasons:

1. An incident-light meter can have a flat rather than hemispherical reflector.
2. The function of the hemispherical receptor on a meter isn't quite the same as that of an integrating sphere, and the link serves more to mislead than educate.

The function of the receptor isn't important in this context; what matter is the incident-light measurement's independence of the reflectance of scene elements. The function of the hemispherical receptor is adequately (and accurately) explained in the section Calibration constants.

I'm going to remove this sentence unless someone has a good reason for keeping it. JeffConrad (talk) 04:52, 9 February 2009 (UTC)

Edit : Discussion moved to "angle of acceptance" topic - Redbobblehat (talk) 20:29, 9 February 2009 (UTC)
My point was that reference to an “integrating sphere” is at best superfluous. Norwood's description of the hemispherical receptor is the best that I have found, but it obviously isn't a solid mathematical treatment. If a more rigorous treatment exists, it would be an excellent addition to the article. JeffConrad (talk) 15:55, 9 February 2009 (UTC)
I accept that “integrating sphere” may not be the best analogy. In the interests of 'preserving the phenomena' would something like "a diffuser with a 180° angle of acceptance" be a better substitute ? It strikes me that "180°" is not strictly accurate, but "2π steradian" is inappropriate to the idiom ... would "hemispherical angle of acceptance" be intelligible enough ? or "a diffuser with a hemispherical field of view" ? --Redbobblehat (talk) 20:29, 9 February 2009 (UTC)

## Angle of Acceptance

Perhaps a section headed Angle of acceptance would be a useful starting point for comparing the practical (and mathematical? ;) differences between "dome" & "flat" incident meters, and "spot" and "matrix average" (or whatever) reflected meters ? I'm thinking that even if angle of acceptance isn't the only factor involved, it could be a helpful 'backbone' for structuring the other factors around; a good place to start for the uninitiated reader ? --Redbobblehat (talk) 14:45, 9 February 2009 (UTC)

Angle of acceptance is applicable to reflected-light meters (and is discussed in ISO 2720), but it's not really applicable to incident-light meters, for which the angle of acceptance is essentially 180° with either a flat or hemispherical receptor—the two differ in the weighting given to off-axis values. Multi-segment (“matrix”, “evaluative”) metering is proprietary to each manufacturer, and consequently tough to cover in this article. And I'm not sure angle of acceptance is applicable. JeffConrad (talk) 15:55, 9 February 2009 (UTC)
It seems to me that any angle of acceptance less than 360° (spherical) is significant for correct use of the meter; as it determines what direction you need point it in to achieve a relevant reading. This seems to be more of an issue for flat/ domed/ conical/ spherical diffusers on Incident meters because they are less intuitive (camera-lens-like) than Reflected meters - should you point it at the camera? or light source? or perpendicular to the subject surface?, etc ... I realise this may seem too basic, but I was hoping to suggest we structure the article around some "basic principles" of light meters ... Hmm. Perhaps angle of acceptance isn't one of them ? --Redbobblehat (talk) 20:29, 9 February 2009 (UTC

## Who intented it?

I have learned that Thomas Edison, the inventor of photography, invented the light meter too. Unfortunately, I've got no reliable references, can anyone help? --141.91.129.2 (talk) 09:32, 21 August 2009 (UTC)

There are many reliable sources that can clear up your idea that he invented photography! Dicklyon (talk) 00:25, 22 August 2009 (UTC)
It is commonly known that Edison invented it. My parents told me so, do you suggest they were liars???? —Preceding unsigned comment added by 84.141.34.227 (talk) 08:11, 23 August 2009 (UTC)

## Light meter equation

The units don't work out on the equation for a light meter . . . this is strange i am curious what the real formula is 128.197.81.26 (talk) 00:01, 16 January 2010 (UTC)

The units work out if you include units with the calibration constants K and C, but ISO standards, like the ANSI and ASA standards before them, tend to ignore such details. This could lead to confusion when working with different unit systems. The current standard, ISO 2720:1974, assumes SI units, but ANSI PH3.49-1971 provided for both cd · m−2 and fL; the constant K was treated as dimensionless, with different numerical values given for the different unit systems.
The equations given are the “real” ones; they don't derive from any physical law—they're just notational conveniences, with the constants somewhat arbitrarily chosen to indicate camera settings that lead to the “best” picture. JeffConrad (talk) 03:28, 16 January 2010 (UTC)

## Confusion between Mid Reflectance Value vis 18% Mid Tone (geometric) Value

There remains perpetual confusion on many photography blogs regarding the above. I think the section Light_Meter#Calibrated_reflectance answers it in part (and yet they still argue). I think it might be easier for lay readers to get their heads around it if we explain that by converting the image into a jpeg or tiff - the noise is simply removed - and this creates a pseudo Ev reduction. Thus making it appear that the light-meter (either built-in or hand-held) is calibrated to ≈18%. The footnote on page 11 of The ISO Definition of the Dynamic Range of a Digital Still Camera, Douglas Kerr makes this easy to understand when explaining noise removal– without the math. Quote:

“It is this margin provision that causes the confusion between the roles of an 18% average reflectance and a 13% average scene reflectance as premises for the “standard” exposure meter calibration; the luminance that is 18% of “70.1% of maximum recordable luminance” is 13% of the maximum recordable luminance itself.”


However, this Wikipedia article is about light meters only and not cameras and what happens to the RAW image data. So I'm not sure how to add this information to the article. I have just looked at the Sekonic L-208 TwinMate specs on their site. They quote: C = 340 and K = 12·5. So, (π*12.5)/340 = 0·1155 or 11·6% if you like. Well within the ISO specifications. Yet converted a RAW file exposed with the help of this instrument to jpeg or tiff and a pseudo change in mid tone will appear as if the camera was calibrated to about 18% (give or take 1/3 of a stop (as allowed in the ISO std for jpegs and tiffs). Have also found a sketch of how to use a grey card properly. [15]. One is aiming for diffuse reflection not specular reflection (which the often quoted angle of 45° will encourage). Also, the card will appear about 1/3 stop dimmer. That together with the ≈ -1/3 stop most digital cameras appear to meter, from a grey card full-on and the Ev will be about right for most shots. I'm hopeless at drawing but if you think it would be useful here and on Grey card I will issue a request on WP for a drawing to be done. After all, the idea itself is not copyrightable although this image is.--Aspro (talk) 02:25, 6 December 2014 (UTC)