Lumens and the Luminosity Function

In summary, the conversation discusses the formula for calculating luminous flux and the role of the luminosity function in correcting for the human eye's sensitivity to different wavelengths. There is confusion about the units and role of the luminosity function, with one person mistakenly believing that it represents a constant amount of watts per nanometer. It is clarified that the luminosity function is a function of wavelength and not a constant. It is also noted that the luminosity function is used in an integral to calculate the total luminous flux of a light source. There is also discussion about the spectrum of a light source and how the luminosity function is not a cumulative sum, but rather a function that takes into account the specific wattage and sensitivity of each wavelength emitted
  • #1
JohnnyGui
796
51
Hello,

I have a small confusion regarding the formula that calculates the luminous flux of a light source, which is given by:

df6223259e88c1cf7e8765bb75ffd570e0c697b8.png

(See this Wiki)

The ##\bar y (λ)## is the so-called luminosity function that corrects for the sensitivity of the human eye for a particular chosen wavelength.
The ##Φ_{e,\lambda}## is the "objective" radiant flux (in watts) per nanometer which is the total power per nanometer, regardless of the sensitivity of the human eye (hence "objective"). You can see from the formula that if a light source emits more than 1 wavelength, an integration has to be done with the limits being the endpoints of the emitting wavelength range, so that it gives the total luminous flux of that light source.

Now, I understand that this formula gives the luminous flux ##Φ_V## in lumen because of the multiplication with ##683.002 lm/W##. However, if we remove this constant, then this means that:

Watts.jpg


Will give a luminous flux in Watts, right? This means that you can calculate the subjective power according to the human eye's sensitivity in Watts. According to the definition of 1 candela, if we now plug in a wavelength of 555nm of 1 Watt in the formula, we'd get 1/683 Watts out of this formula. This means that the subjective power of 1 candle according to the human eye is 1/683 Watts.

If that's the case, then I don't get why this Wiki says the following:

"The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540×1012 hertz and that has a radiant intensity in that direction of  1⁄683 watt per steradian"

Notice that it says that 1 candela is the radiant intensity of 1/683 watt, which is the overall objective power (per steradian in this case) and not the luminous intensity (per steradian) as I concluded, which is the subjective power according to the human eye.

What am I concluding wrong here?
 
Physics news on Phys.org
  • #2
Ok, so I’ve figured out what my misunderstanding was. The Luminous Flux (the formula above) merely shows how strong a light source is relative to a candle that emits light of 555nm at 1/683 Watts, all based on the human eye’s sensitivity.

I have one other question. I understand that if a light source emits several wavelengths, that you’d have to integrate the formula above over dλ as discussed. For each dλ you’d have to give the specified amount of Watts that the light source is emitting for that specified dλ.

What I don’t understand is that this is given by a constant ##Φ_{e,λ}## that gives the amount of Watts per nanometer in the luminosity function. This constant says that the larger the wavelength, the larger the amount of Watts it emits and that it's proportional. But a light source doesn’t necessarily emit larger wavelengths at larger Watts and it surely doesn't emit wavelengths that is proportional to the amount of Watts. So why is this constant in the luminosity function?
 
  • #3
You are assuming a spectrum in which power increases with wavelength. That is almost never the case. When you restrict the range of interest to the visible spectrum (360nm to 830nm) then that is case for a Planckian or black body emitter for temperatures below about 3500°K. According to Wein's Displacement Law, the peak wavelength of a Planckian emitter scales as 1/absolute temperature (http://hyperphysics.phy-astr.gsu.edu/hbase/wien.html).

Even allowing for that, the luminosity function has a maximum at 555nm and drops rapidly as you move to longer and shorter wavelengths. So at 555nm it is 1.0000, at 600nm it is 0.6292, at 650nm 0.1069, at 700nm 0.0041, at 750nm 0.0001 and it keeps dropping.

For your statement to be true, the power at each wavelength would have to increase faster than the luminosity function is decreasing. Anything that does that would appear as a very pure red color.
 
  • #4
Eric Bretschneider said:
You are assuming a spectrum in which power increases with wavelength. That is almost never the case.

But I agree on that power shouldn't increase with wavelength. My problem is that the formula of the luminosity function shows a constant of Watts/nm which implies that the amount of Watts increases with wavelength My post was directed at the fact that this shouldn't be the case.

This Wiki gives the following formula for the luminosity function:

df6223259e88c1cf7e8765bb75ffd570e0c697b8.png

In which ##Φ_{eλ}## is the spectral radiant flux, in watts per nanometer. How can this be while the wavelength is never proportional to the amount of watts in the spectrum of a lightsource (which is a peak curve)?
 
  • #5
Watts/nm isn't a constant - it is the units!

Φ is a function of the wavelength. Don't confuse units with numbers.
 
  • #6
Eric Bretschneider said:
Watts/nm isn't a constant - it is the units!

Φ is a function of the wavelength. Don't confuse units with numbers.

Ah, I got confused there! That's why it has a ##(λ)## after it since it differs with the wavelength.

So is ##Φ_{eλ}## the derivative of a spectrum curve of a light source so that integrating it over a specified spectrum range would give the total energy in that spectrum range in Watts?
 
  • #7
I am kind of surprised that the luminosity function needs to be an integral if a light source is emitting a spectrum of wavelengths.

From what I understand, each specific wavelength which is emitted, has a certain amount of watts and a certian sensitivity of ##\overline{y}(λ)##. So it's actually not the amount of watts and sensitivity in a given range of ##dλ##, which is what the integral does, but for each specific wavelength.
So instead, I'd expect that it has to be a cumulative sum that adds every amount of watts and sensitivity of each specific wavelength over the range of the spectrum of a light source.

Shouldn't therefore the luminosity function be a summation formula that adds ##Watts(λ) \cdot \overline{y}(λ)## for each wavelength that the light source emits?
 
  • #8
The sensitivity function is also known as V(λ) and has units of lumens/W. So the spectral power distribution (Watts/nm) x V(λ) gives lumens (when you multiple by a wavelength interval).

Integration is a summation process. You can calculate an analytical integral (which would be an equation) or a numerical integral which would be the process you are describing. They are one and the same.
 
  • #9
Eric Bretschneider said:
The sensitivity function is also known as V(λ) and has units of lumens/W. So the spectral power distribution (Watts/nm) x V(λ) gives lumens (when you multiple by a wavelength interval).

Integration is a summation process. You can calculate an analytical integral (which would be an equation) or a numerical integral which would be the process you are describing. They are one and the same.

But doesn't multiplying the spectral power distribution in Watts/nm with a wavelength interval ##dλ## merely give the difference in Watts between 2 wavelengths with a difference of ##dλ##? So that basically integration is merely giving the total difference of Watts between the 2 ends of a spectrum range? Basically like this (which is in this case a total difference of 680nm):

Integration.jpg


What I meant with summation is more like a function with ##∑## so that the formula adds the amount of Watts of each specific wavelength in the spectrum range together. Like this:

Summation.jpg
 
  • #10
No it doesn't. Integration gives the area under the curve. The spectral power distribution has units of W/nm. If you integrate you multiple W/nm by nm and get W. When you take the product of the spectral power distribution and the V(λ) function you get W/nm x lumens/W = lumens/nm. Integrating that gives you lumens.

In numerical integration you would take the value at every wavelength and multiple by the spacing and add it all up. The simplest form of numerical integration is adding the area of a bunch of rectangles that just touch the curve you are integrating (start, mid point or end of the wavelength interval). You could also do trapezoidal integration or something like Simpson's Rule (fits a parabola to 3 data points). Integration is the same thing as summation!

Here is a simplified tutorial https://www.mathsisfun.com/calculus/integration-introduction.html
 
  • #11
Eric Bretschneider said:
No it doesn't. Integration gives the area under the curve. The spectral power distribution has units of W/nm. If you integrate you multiple W/nm by nm and get W. When you take the product of the spectral power distribution and the V(λ) function you get W/nm x lumens/W = lumens/nm. Integrating that gives you lumens.

In numerical integration you would take the value at every wavelength and multiple by the spacing and add it all up. The simplest form of numerical integration is adding the area of a bunch of rectangles that just touch the curve you are integrating (start, mid point or end of the wavelength interval). You could also do trapezoidal integration or something like Simpson's Rule (fits a parabola to 3 data points). Integration is the same thing as summation!

Here is a simplified tutorial https://www.mathsisfun.com/calculus/integration-introduction.html

I think my whole misunderstanding here is that I think that ##Φ_{eλ}(λ)## is a derivative of that spectrum graph I posted in my post #10 above.
Integrating a derivative will give the area beneath its own graph line like you said. However, that area beneath that derivative graph line is equal to the total difference between your 2 chosen limits in the original function (the one you took the derivative of).

Since I thought that ##Φ_{eλ}(λ)## is the derivative function of that spectrum graph, I expected it to merely give the difference in amount of watts between 2 chosen wavelengths in the spectrum, when you integrate it.
 
  • #12
Definitely not. It isn't even an analytical function. Although you can go to some extremes to create a piece-wise analytical version, in practice it is much easier to use a lookup table.
 
  • #13
Eric Bretschneider said:
Definitely not. It isn't even an analytical function. Although you can go to some extremes to create a piece-wise analytical version, in practice it is much easier to use a lookup table.

Apologies for my stubborness but I've found a simpler way to explain what I don't understand.

Suppose a light source is emitting light only between 2 and 5 nm, and the spectral power (Wtts/nm) being constant at 4 Wtts/nm over that range of wavelengths (I know it's not supposed to be constant, but just for the sake of argument). The spectral power graph would then look like this:

Spectral Power graph.jpg

The luminosity function would integrate this graph. Integration will give an area of 12 Watts.

However, this graph does not represent the absolute power in Watts that each wavelength is emitted at. Spectral power is more like the power concentration of each emitted wavelength. And from what I understand, to know the amount of lumen a wavelength has, one has to look at its absolute power that it is emitted at.

The spectral power (power concentration) of 4 Wtts/nm multiplied by the corresponding wavelength will give the absolute power (##P##) for that wavelength. Multiplying that absolute power (in Watts) of each wavelength with the sensitivity function (Lumens/Wtts) will give the total amount of lumens for that specific wavelength. I'd have to do this for each wavelength and add them all together to get the total amounts of lumens over the whole spectrumrange (2nm -5nm)
So the formula would look something like this:

##P(λ=2) \cdot \overline{y}(λ=2) + P(λ=3) \cdot \overline{y}(λ=3) + P(λ=4) \cdot \overline{y}(λ=4) + P(λ=5) \cdot \overline{y}(λ=5)##

This formula however will give way more total amounts of Watts than the 12 Watts calculated in the spectral power graph.

So my confusion lies in the fact that the luminosity formula uses the spectral power of a wavelength x the corresponding sensitivity function to get the amount of lumens at that wavelength while I think that it should take the absolute power of that wavelength and multiply it with the wavelength to get the amount of lumens at that wavelength. This is what I can't seem to grasp.
 
Last edited:
  • #14
You have it correct, but remember that when you multiply by V(λ) the units of the product are no longer Watts. V(λ) has units of lumens/Watt so the number that you get (larger than 12 Watts) is the number of lumens. You should be using the absolute power at a wavelength x V(λ) at that wavelength.

Lumens are a perceptual unit and that makes things peculiar at times. 100 lumens should appear as the same "brightness" regardless of the spectral power distribution.
 
  • #15
Eric Bretschneider said:
You have it correct, but remember that when you multiply by V(λ) the units of the product are no longer Watts. V(λ) has units of lumens/Watt so the number that you get (larger than 12 Watts) is the number of lumens. You should be using the absolute power at a wavelength x V(λ) at that wavelength.

My bad, I indeed meant that I'd get way more lumens than when I integrate the spectral power graph according to: ##Φ_{eλ}(λ) \cdot V(λ)\cdot dλ##.
If I have it correct, why would integration of that luminosity formula then give way less lumens than my conclusion?
 
  • #16
JohnnyGui said:
My bad, I indeed meant that I'd get way more lumens than when I integrate the spectral power graph according to: ##Φ_{eλ}(λ) \cdot V(λ)\cdot dλ##.
If I have it correct, why would integration of that luminosity formula then give way less lumens than my conclusion?
Could someone please help me with this question? Still not really understand why my conclusion is different from how it is actually calculated while it should be the absolute power for each wavelength.
 
  • #17
JohnnyGui said:
Could someone please help me with this question? Still not really understand why my conclusion is different from how it is actually calculated while it should be the absolute power for each wavelength.

Bump
 
  • #18
JohnnyGui said:
Hello,

I have a small confusion regarding the formula that calculates the luminous flux of a light source, which is given by:
<snip> However, if we remove this constant, then this means that:

View attachment 114991

Will give a luminous flux in Watts, right?

So far, so good- the formula above is the power ('light watts') that gives rise to a response by the human eye. The prefactor 683 lm/W at 540 THz is a normalization factor for human vision (hence the Wiki entry). There are other normalization factors: 'Photosynthetic watts', for example, or 'sunburn watts'. 1 sunburn watt = 1 finsen.

These normalization factors (and others- the sidewinder missile uses 'lead sulfide watts') are used to avoid the need for spectral radiometric measurements.

Don't forget- 'intensity' is W/sr.

Does that help?
 
  • #19
Andy Resnick said:
So far, so good- the formula above is the power ('light watts') that gives rise to a response by the human eye. The prefactor 683 lm/W at 540 THz is a normalization factor for human vision (hence the Wiki entry). There are other normalization factors: 'Photosynthetic watts', for example, or 'sunburn watts'. 1 sunburn watt = 1 finsen.

These normalization factors (and others- the sidewinder missile uses 'lead sulfide watts') are used to avoid the need for spectral radiometric measurements.

Don't forget- 'intensity' is W/sr.

Does that help?

Thanks for the reply. I indeed was able to understand this eventually. However, a new confusion came regarding calculating the spectral sensitivity. I can't formulate my question better than I have in my post #14. Hope you're able to help me with that
 
  • #20
JohnnyGui said:
Thanks for the reply. I indeed was able to understand this eventually. However, a new confusion came regarding calculating the spectral sensitivity. I can't formulate my question better than I have in my post #14. Hope you're able to help me with that

Not exactly sure what you mean: liberal snipping has occurred...

"Suppose a light source is emitting light only between 2 and 5 nm, and the spectral power (W/nm) being constant at 4 W/nm over that range of wavelengths
The luminosity function would integrate this graph. Integration will give an area of 12 W.

However, this graph does not represent the absolute power in Watts that each wavelength is emitted at. Spectral power is more like the power concentration of each emitted wavelength. And from what I understand, to know the amount of lumen a wavelength has, one has to look at its absolute power that it is emitted at.

So my confusion lies in the fact that the luminosity formula uses the spectral power of a wavelength x the corresponding sensitivity function to get the amount of lumens at that wavelength while I think that it should take the absolute power of that wavelength and multiply it with the wavelength to get the amount of lumens at that wavelength. This is what I can't seem to grasp."

The first paragraph is fine- only you don't need a 'luminosity function'. You already specified the spectral power output. In your second paragraph, you *already* specified the spectral power output: constant at 4W/nm. Over a 1 nm waveband, you have 4 W. Over a 0.1 nm waveband, you have 0.4 W. Etc., etc.

I wonder if you are hung up on continuum formulas (∫ (stuff)dλ) versus discrete: output power at 543.87684 nm.

Not sure what you are struggling with...
 
  • #21
Andy Resnick said:
Not exactly sure what you mean: liberal snipping has occurred...

"Suppose a light source is emitting light only between 2 and 5 nm, and the spectral power (W/nm) being constant at 4 W/nm over that range of wavelengths
The luminosity function would integrate this graph. Integration will give an area of 12 W.

However, this graph does not represent the absolute power in Watts that each wavelength is emitted at. Spectral power is more like the power concentration of each emitted wavelength. And from what I understand, to know the amount of lumen a wavelength has, one has to look at its absolute power that it is emitted at.

So my confusion lies in the fact that the luminosity formula uses the spectral power of a wavelength x the corresponding sensitivity function to get the amount of lumens at that wavelength while I think that it should take the absolute power of that wavelength and multiply it with the wavelength to get the amount of lumens at that wavelength. This is what I can't seem to grasp."

The first paragraph is fine- only you don't need a 'luminosity function'. You already specified the spectral power output. In your second paragraph, you *already* specified the spectral power output: constant at 4W/nm. Over a 1 nm waveband, you have 4 W. Over a 0.1 nm waveband, you have 0.4 W. Etc., etc.

I wonder if you are hung up on continuum formulas (∫ (stuff)dλ) versus discrete: output power at 543.87684 nm.

Not sure what you are struggling with...

Ok, let's take this step by step to see if I understand this. Suppose a light source is emitting just one exact wavelength, which is 500 nm, at 2 Watts. The sensitivity ##\overline{y}(λ)## of the human eye for any wavelength at a particular amount of Watts is measured with respect to a wavelength of 555nm at 1/683 Watts. When looking at the graph for eye sensitivity, I can read that the ##\overline{y}(500nm) = 0.3##.
This means that the eye sensitivity of a wavelength of 500nm is 30 percent of the sensitivity of 555nm if both wavelengths are emitted with the same amount of Watts. The more Watts a wavelength is emitted at, the more the eye is sensitive to it.
This means that a wavelength of 500nm at 2 Watts would be ##0.3 \cdot \frac{200}{\frac{1}{683}} = 409.8## times more sensitive than a wavelength of 555nm at 1/683 Watts, thus 409.8 Lumens.

This is done without using the luminosity function. Let's now see how it's calculated using the luminosity function, which says that:
$$Φ_{v} = 683 Lmn / W \cdot \overline{y}(λ) \cdot Φ_{eλ} \cdot \lambda$$
The units of ##Φ_{eλ}## is Watts/nm, so that means that ##Φ_{eλ}## for a wavelength of 500nm at 2 Watts is ##Φ_{eλ}= \frac{2 Watts}{500nm} = 0.004W/nm##. Filling the data in the formula, ##\overline{y}(500nm) = 0.3##, ##\lambda = 500nm## and ##Φ_{eλ}= 0.004W/nm##, indeed gives the same amount of Lumens as before: 409.8 Lumens.

Before I start with my struggle, is what I typed correct so far?
 
  • #22
JohnnyGui said:
Ok, let's take this step by step to see if I understand this. Suppose a light source is emitting just one exact wavelength, which is 500 nm, at 2 Watts.

Ok, stop there. Do you mean the source emits 2W at 500.0000000... ±0 nm? That's fine- but then what is your spectral power output? (hint- it's a delta function).

JohnnyGui said:
The sensitivity ##\overline{y}(λ)## of the human eye for any wavelength at a particular amount of Watts is measured with respect to a wavelength of 555nm at 1/683 Watts. When looking at the graph for eye sensitivity, I can read that the ##\overline{y}(500nm) = 0.3##.
This means that the eye sensitivity of a wavelength of 500nm is 30 percent of the sensitivity of 555nm if both wavelengths are emitted with the same amount of Watts. The more Watts a wavelength is emitted at, the more the eye is sensitive to it.
This means that a wavelength of 500nm at 2 Watts would be ##0.3 \cdot \frac{200}{\frac{1}{683}} = 409.8## times more sensitive than a wavelength of 555nm at 1/683 Watts, thus 409.8 Lumens.

Where did that factor of 200 come from? Regardless, when I did the calculation, I also got that a 2W source at 500nm is 409.8 times as intense as a 1/683 W source at 555 nm.

JohnnyGui said:
This is done without using the luminosity function. Let's now see how it's calculated using the luminosity function, which says that:
$$Φ_{v} = 683 Lmn / W \cdot \overline{y}(λ) \cdot Φ_{eλ} \cdot \lambda$$
The units of ##Φ_{eλ}## is Watts/nm, so that means that ##Φ_{eλ}## for a wavelength of 500nm at 2 Watts is ##Φ_{eλ}= \frac{2 Watts}{500nm} = 0.004W/nm##.

Here's two errors- $$Φ_{v} = 683 Lmn / W \cdot ∫\overline{y}(λ) \cdot Φ_{eλ} \cdot d\lambda$$ and ##Φ_{eλ}= (2 W) δ(500 nm - λ)##

Evaluating the integral gives 409.8 lm as before.

Does this help?
 
  • #23
Andy Resnick said:
Ok, stop there. Do you mean the source emits 2W at 500.0000000... ±0 nm? That's fine- but then what is your spectral power output? (hint- it's a delta function).

Yes, that's what I mean. I take the spectral power output is in Watts/nm? What I did here is say that 2 Watts is the total power that the 500nm is being emitted at, so I kind of did it the other way round and thus calculate that the spectral power output must be 2W / 500nm = 0.004W/nm.

Andy Resnick said:
Where did that factor of 200 come from? Regardless, when I did the calculation, I also got that a 2W source at 500nm is 409.8 times as intense as a 1/683 W source at 555 nm.

Apologies! That was a typo and I meant to type 2, I'll fix it in the post.

Andy Resnick said:
Here's two errors-
Φv=683Lmn/W⋅∫¯¯¯y(λ)⋅Φeλ⋅dλΦv=683Lmn/W⋅∫y¯(λ)⋅Φeλ⋅dλ​
Φ_{v} = 683 Lmn / W \cdot ∫\overline{y}(λ) \cdot Φ_{eλ} \cdot d\lambda and Φeλ=(2W)δ(500nm−λ)Φeλ=(2W)δ(500nm−λ)Φ_{eλ}= (2 W) δ(500 nm - λ)

I threw out the integration symbol because here I'm merely caculating just 1 wavelength, so an integration is not needed as far as I understand.

I don't really get the equation: ##Φ_{eλ}= (2 W) δ(500 nm - λ)##. What is the amount of ##\lambda## and why does it have to be substracted from 500nm?
 
  • #24
JohnnyGui said:
Yes, that's what I mean. I take the spectral power output is in Watts/nm? What I did here is say that 2 Watts is the total power that the 500nm is being emitted at, so I kind of did it the other way round and thus calculate that the spectral power output must be 2W / 500nm = 0.004W/nm.

But that's not right. That's not what 'spectral power output' means.
JohnnyGui said:
I threw out the integration symbol because here I'm merely caculating just 1 wavelength, so an integration is not needed as far as I understand.
I don't really get the equation: ##Φ_{eλ}= (2 W) δ(500 nm - λ)##. What is the amount of ##\lambda## and why does it have to be substracted from 500nm?

The symbol δ(λ0-λ) refers to a 'Dirac delta function'. It's value is zero when λ≠λ0 and ∞ when λ=λ0. It has a normalization property such that ∫δ(λ0-λ) dλ = 1. Similarly, ∫F(λ)δ(λ0-λ) dλ = F(λ0).

Does that help?
 
  • #25
Andy Resnick said:
But that's not right. That's not what 'spectral power output' means.

Ah, then that's probably what the cause of my confusion was regarding the integration.

Andy Resnick said:
The symbol δ(λ0-λ) refers to a 'Dirac delta function'. It's value is zero when λ≠λ0 and ∞ when λ=λ0. It has a normalization property such that ∫δ(λ0-λ) dλ = 1. Similarly, ∫F(λ)δ(λ0-λ) dλ = F(λ0).

Went to read the Wiki about the Dirac delta function but not really getting the full grasp of it. Some questions on this if you don't mind:
1. I take it the λ0 is the emitted wavelength and λ the 555 nm?
2. If if a light source only emits one wavelength, like my previous example of 500nm, does that mean that ##Φ_{eλ} = 1##?
3. How is this Dirac delta unit expressed and calculated when a lightsource emits a range of wavelengths, for example from 500 nm to 525 nm? Really would appreciate an example regarding this.
I myself, if I read that a light source is emitting a range of wavelengths, for example λ1 up to λ10, then I'd conclude that the total amount of Lumens is:

$$ Lumens = \overline{y}(\lambda_1) \cdot Watts(\lambda_1) \cdot 683 W/Lmn + \overline{y}(\lambda_2) \cdot Watts(\lambda_2) \cdot 683 W/Lmn + \overline{y}(\lambda_3) \cdot Watts(\lambda_3) \cdot 683 W/Lmn + ... +\overline{y}(\lambda_{10}) \cdot Watts (\lambda_{10}) \cdot 683 W/Lmn$$
I'm curious using this dirac delta ##Φ_{eλ}## would give the same answer as this formula or not.
 
Last edited:
  • #26
JohnnyGui said:
Some questions on this if you don't mind:
1. I take it the λ0 is the emitted wavelength and λ the 555 nm?

No- if your source emits 2W at 500.00... ±0 nm, the spectral power output is 2 δ(500-λ) W.

JohnnyGui said:
2. If if a light source only emits one wavelength, like my previous example of 500nm, does that mean that ##Φ_{eλ} = 1##?

No, it means ##Φ_{eλ} = 2 δ(500-λ) W##.

JohnnyGui said:
3. How is this Dirac delta unit expressed and calculated when a lightsource emits a range of wavelengths, for example from 500 nm to 525 nm? Really would appreciate an example regarding this.

There are at least two ways to do this. One is to specify the spect5ral distribution- say, for example, the source emits a Gaussian spectral distribution with an average wavelength of λ = 525 nm and a FWHM of 20 nm. The spectral power output would be something like ##Φ_{eλ} = P_0 exp(-([λ-525]/10)^2)##. If, on the other hand, your source emits 1 W at 500.000...±0nm and 0.5 W at 501.0000...±0nm and 0.5W at 453.4367000... ±0nm, ##Φ_{eλ} = δ(500-λ)+0.5 δ(501-λ)+0.5 δ(453.4367-λ)##. One may write the Gaussian source as an integral of Delta-function sources, but it's rather cumbersome and usually there is little to be gained.

If you have some arbitrary spectral power output ##Φ_{eλ}## and the total power output is 2W, then ##Φ_{v}=P_0∫Φ_{eλ}dλ##, where P_0 =2/∫Φ_{eλ}dλ.

Does this help?
 

Related to Lumens and the Luminosity Function

1. What is a lumen?

A lumen is a unit of measurement for the total amount of visible light emitted by a light source. It is a measure of the total amount of energy that is perceived by the human eye as brightness.

2. How is the luminosity of a light source measured?

The luminosity of a light source is measured in lumens (lm). This is done by using a device called a photometer, which measures the total amount of visible light emitted by the source in all directions.

3. What is the Luminosity Function?

The Luminosity Function is a mathematical function that represents the sensitivity of the human eye to different wavelengths of light. It takes into account the fact that the human eye is more sensitive to certain colors than others, and therefore, not all wavelengths of light are perceived as equally bright.

4. How does the Luminosity Function affect the perceived brightness of a light source?

The Luminosity Function plays a crucial role in determining the perceived brightness of a light source. It shows that the human eye is most sensitive to green light and least sensitive to red and blue light. This means that a light source with a higher proportion of green light will appear brighter to the human eye, even if it emits the same number of lumens as a light source with a different color balance.

5. Why is understanding lumens and the Luminosity Function important?

Understanding lumens and the Luminosity Function is important for several reasons. Firstly, it allows us to accurately measure and compare the brightness of different light sources. Secondly, it helps in designing lighting systems that are energy-efficient and provide the desired level of brightness. Lastly, it is crucial in the development of technologies such as LED lights, which have different color balances and therefore, require a better understanding of the Luminosity Function for optimal performance.

Similar threads

  • Other Physics Topics
Replies
2
Views
3K
Replies
8
Views
3K
  • Classical Physics
Replies
10
Views
1K
Replies
6
Views
1K
  • Other Physics Topics
Replies
10
Views
7K
Replies
2
Views
1K
Replies
2
Views
1K
  • Other Physics Topics
Replies
13
Views
6K
Replies
1
Views
1K
  • Other Physics Topics
Replies
4
Views
14K
Back
Top