Talk:Luminous intensity

Circular

Is it just me, or does this cluster of articles define lumen in termas of candela and candela in terms of lumen?—The preceding unsigned comment was added by 206.165.217.125 (talk • contribs) .

It's just you. "One candela is defined as the luminous intensity of a monochromatic light source of frequency 540 THz...which has a radiant intensity of 1/683 watts per steradian." I'll change "which" to "that" to make it sensible. Dicklyon 22:42, 31 July 2006 (UTC)

But doesn't that still define intensity in terms of intensity? Larklight (talk) 19:23, 14 May 2008 (UTC)

No. Luminous intensity and radiant intensity are two different things. It is not defining one in terms of the other anyway. It is an operational definition. What it does is it describes physically how to make a light source that you know emits one candela of luminous intensity. The candela is an SI base unit, so it is defined by experiment rather than directly referenced to other SI base units. The definition says that if you take a light source that emits monochromatic 540 THz light, and adjust it so its radiant intensity is 1/683 watts per steradian, the source's luminous intensity will be one candela. --Srleffler (talk) 22:51, 14 May 2008 (UTC)

Do we need Luminous Intensity?

The candela is the only SI unit that is intended to map human perception. We don't have an SI unit for sound, or flavor. Why do we need one for light? Dr Thermo (talk) 02:58, 12 July 2011 (UTC)

It's partly historical, that at the time when the system was established, we did not have a way to measure luminous intensity in terms of the other base units. At the time, it could be measured only by having a human compare the intensity of a source to a reference of known luminous intensity. Improvements in radiometry and characterization of the human eye's response (see Luminosity function) made it possible to define the candela in terms of other units. If we were creating it today, I imagine it would be a derived unit, but for historical reasons it has been retained as a "base unit".

I suppose we don't have a specific SI unit for sound intensity because in practice standard units of power are sufficient. For flavor, I imagine the reason is a combination of lack of ability to quantify it and lack of applications for a quantitative measurement of it.--Srleffler (talk) 04:02, 12 July 2011 (UTC)

SI

Why is this an 'SI base unit' ? Perhaps the article could explain.—The preceding unsigned comment was added by 70.225.166.87 (talk • contribs) 20:11, October 19, 2006.

The SI base units were chosen to give a complete set of physical quantities that can be precisely measured, so that the definitions of the units are precise. The explanation belongs at SI base unit, and this article does provide a link there.--Srleffler 02:19, 20 October 2006 (UTC)

One of the things that distinguishes the base units from other units is that their definitions are operational. The base units are not directly defined in terms of other units, but rather are defined by a description of a physical process that would produce exactly one unit of the specified quantity. --Srleffler (talk) 23:18, 14 May 2008 (UTC)

Symbol

In the right side bar, the symbol for luminous intensity is given as uppercase L but in calculations later in the article, uppercase I is used. I believe uppercase I is correct but would appreciate confirmation. ErmBee (talk) 12:14, 29 October 2018 (UTC)

Yep.--Srleffler (talk) 03:21, 30 October 2018 (UTC)

Integral in Usage section

I removed the note reading "!-- TBD: Is integral from 0 to infinity really correct, or should it better be 380nm to 780nm here? --". The answer is that, strictly speaking, 0 to infinity is correct, but 380 to 780 nm is close enough, since the integrand has *experimentally* been measured to be zero, to within experimental error, outside of the 380-780 nm interval. If the $y(\lambda )$ is ever measured extremely accurately, the 0 to infinity definition will remain valid, while the 380-780 nm may not be. Even though it may never be measured that accurately, it's just bad practice.

In addition, I have a question about that integral:

$$I_{\mathrm {v} }=683\int _{0}^{\infty }{\overline {y}}(\lambda )\cdot {\frac {dI_{\mathrm {e} }(\lambda )}{d\lambda }}\,d\lambda .$$

I cannot make sense of this, I would expect: $I_{\mathrm {v} }=683\int _{0}^{\infty }{\overline {y}}(\lambda )\cdot I_{\mathrm {e} }(\lambda )\,d\lambda .$

If I have an equal-energy spectrum $I_{\mathrm {e} }(\lambda )=1$, then the first integral is zero, which is not correct. PAR (talk) 03:06, 27 March 2024 (UTC)

I think it's right; the notation gets a bit confusing. Spectral intensity is given by ${\textstyle I_{\mathrm {e} ,\Omega ,\lambda }={\frac {\partial I_{\mathrm {e} ,\Omega }}{\partial \lambda }}}$. The integral is over the product of the spectral intensity times the luminous efficiency function. If you're imagining a flat spectrum it is ${\textstyle I_{\mathrm {e} ,\Omega ,\lambda }}$ that is constant, not ${\textstyle I_{\mathrm {e} ,\Omega }}$

The radiant intensity ${\textstyle I_{\mathrm {e} ,\Omega }}$ is not a function of $\lambda$ in the usual way; it is the radiant intensity of all the light present, at all wavelengths. To apply the luminous efficiency function you have to split off the slice of intensity that depends on each wavelength, which is done by the partial derivative. --Srleffler (talk) 19:05, 29 March 2024 (UTC)