# Brightness & Efficiency calculations
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**Caution!**\
These calculations cannot take many critical parameters into account (such as FP expression level, photobleaching, illumination power, filter delamination, etc... )! As such, they are merely intended as theoretical predictions, and may not reflect the actual comparative brightness of two fluorophores on your system.
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The first number in this field (11.71 below) provides a rough estimate of the apparent [brightness](#brightness) for a given fluorophore/filter-set combination (when [extinction coefficient](https://help.fpbase.org/glossary#extinction-coefficient) and [quantum yield](https://help.fpbase.org/glossary#quantum-yield) are available for the fluorophore). The numbers in parentheses give the [excitation efficiency](#excitation-efficiency) and [collection efficiency](#emission-efficiency) with the current filter combination.
## Brightness
Brightness is calculated as the product of the excitation and collection efficiencies (described below) and the extinction coefficient and quantum yield of the selected fluorophore, all divided by 1000. If the EC and QY are not available for a given probe, then only excitation and collection efficiencies will be shown. The absolute value of this number is not particularly meaningful, but it can be used to **compare** the relative brightness of different fluorophore/filter arrangements.
## Excitation Efficiency
#### "Standard" Mode:
In the normal mode, excitation efficiency is interpreted as *the percentage of light incident upon the sample that can be absorbed by the fluorophore*. It is calculated as the area under the curve for the combined light + excitation filters (and dichroics) + fluorophore excitation spectra, divided by the area under the curve of the light + excitation filters alone:
$$
\frac{ \int ( \epsilon\_{ex} \times fluor\_{ex})}{\int \epsilon\_{ex} }
$$
where $$\epsilon\_{ex}$$ is the effective excitation spectrum: $$\epsilon\_{ex} = light \times filter\_{ex} \times filter\_{dichroic}$$
For example, a (narrow band) laser at the peak absorption wavelength of a fluorophore would have near 100% efficiency; but a very broadband excitation spectrum, even if it overlaps the peak absorption wavelength, can have relatively poor excitation efficiency if it contains excess off-peak energy. Even though the 460/80x filter shown in the first image below covers much of the EGFP excitation spectrum, it has lower excitation efficiency (58%, represented by the area with diagonal lines) than a 488nm laser right at the peak excitation wavelength of EGFP (99.8% efficiency):
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This mode really favors light sources that **efficiently** excite the fluorophore, without a lot of off-peak excitation (which would be particularly useful in a live-cell setting, where you want to minimize the energy impinging on the sample). One downside of this mode is that *broader-band* excitation filters tend to render a **lower** efficiency (and therefore "brightness") score than *narrow-band* filters centered on the peak excitation wavelength... which can be a bit counter-intuitive. Which is why "[broadband mode](#broadband-mode)" is also available, below.
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Another potential confusion here is caused by the fact that we cannot make any assumptions about the **power** of the light source (it is assumed to be "sufficiently high"). As an admittedly "strange" example: a laser at peak absorption wavelength (here, 488nm) will *still* have excellent excitation efficiency even if a poorly matched excitation filter that effectively blocks the laser (eg. 515/20) is added to the light path (since one could theoretically just turn up the laser infinitely).
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#### "Broadband" mode
Because the standard mode of excitation efficiency penalizes broadband "off-peak" excitation, it can lead to unexpected results when comparing the expected brightness of two excitation filters. In "broadband" mode, the light source is assumed to be of constant power, and excitation efficiency is interpreted as *coverage of the excitation spectrum.* Here, efficiency is calculated as the area under the curve for the combined light + excitation filters (and dichroics) + fluorophore excitation spectra, divided by the area under the curve of fluorophore excitation spectra alone:
$$
\frac{ \int ( \epsilon\_{ex} \times fluor\_{ex})}{\int fluor\_{ex}}
$$
where again: $$\epsilon\_{ex}$$ is the effective excitation spectrum: $$\epsilon\_{ex} = light \times filter\_{ex} \times filter\_{dichroic}$$
If you've got a broadband light-source (such as a metal-halide or multi-LED light source) and you are trying to determine the expected brightness of a fluorophore given different excitation filters, this mode is more likely to behave as you would expect.
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The problem with this mode is that it makes laser illumination look *terrible:* as a monochromatic light-source, will never cover much of the fluorophore excitation spectrum, but rather puts a lot of power into (hopefully) the most effective wavelengths for excitation. For this reason, the "[standard mode](#standard-excitation-efficiency)" above is default.
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## Emission (Collection) Efficiency
Excitation efficiency *is the percentage of emission photons that can be collected* given the emission path. It is calculated as the area under the curve for the combined emission filters (and dichroics) + camera QE + fluorophore emission spectra, divided by the area under the full fluorophore emission spectrum.
$$
\frac{ \int ( \epsilon\_{em} \times fluor\_{em})}{\int fluor\_{em}}
$$
where $$\epsilon\_{em}$$ is the combined emission path spectrum: $$\epsilon\_{em} = filter\_{em} \times filter\_{dichroic} \times camera\_{QE}$$
In the image below, the EGFP emission spectrum is relatively well matched to the 525/50m filter, and the collection efficiency, represented by the area with diagonal lines, is about 58% the area of the full fluorophore emission spectrum.
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As a side-note: don't forget that many additional photons are lost as a result of not being collected by the objective lens in the first place, or by scattering or absorption somewhere in the emission path. So "100%" collection efficiency here by no means that you collected every photon emitted from the fluorophore.
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