A ‘one channel’spectral colour prediction model for transparent fluorescent inks on a...

transcript

* Proceedings of the IS&T/SID 5

Color Imaging Conference: Color Science, Systems and Applications, November 17-20, 1997,Scottsdale, Arizona, USA, pp. 70-77.

A “one channel” Spectral Colour Prediction Model for Transparent Fluorescent Inks on a

Transparent Support

Patrick Emmel, Roger David HerschLaboratoire de Systèmes Périphériques

Ecole Polytechnique Fédérale de Lausanne (EPFL),CH-1015 Lausanne, Switzerland.

emmel@di.epfl.ch, hersch@di.epfl.ch

1. Introduction

Classical colour prediction models do not take fluorescenceinto account, and their predictions fail if fluorescentsubstances are contained in the inks. This is due to the factthat the overall spectrum of fluorescent substances dependson the spectrum of the light source. Furthermore,fluorescence is a non linear phenomenon. This study aims atpredicting the spectra and the colours of uniform samplesproduced with one fluorescent ink at different concentrations.The method is based on a “one channel” modelization of thephenomenon where only one direction of propagation istaken into account. This model is used to compute thetransmittance spectra of uniform colour samples. Theproposed prediction model requires measuring thetransmittance spectra, the quantum yields, the absorptionbands and the emission spectra of the fluorescent inks. Incontrast to existing fluorescence prediction methods,

3,9,13

our approach enables, without additional measurements, topredict spectra for different ink concentrations and differentlight sources. Moreover, the mathematical formalism wehave developed can be seen as a generalization of Beer’sabsorption law. We hereby obtain accurate spectralpredictions of colour patches. In order to limit the number ofphysical parameters required, the study is carried out withtransparent inks on a transparent substrate, thus avoiding allproblems related to light diffusion.

2. Bouguer-Lambert-Beer law

Let us consider a light-absorbing medium of infinitesimalthickness (see Fig. 1). According to Bouguer,

theintensity attenuation of a monochromatic light passingthrough the medium is proportional to the light intensity and to the thickness , hence:

dφ λ x,( ) m λ( ) φ λ x,( )dx–=

The factor is the

natural absorbance index

of themedium in question. Lambert showed that for liquids and gasthis factor is proportional to the concentration of theabsorbing substance:

The function is called

decadic extinctioncoefficient

. We therefore have:

By integrating equation (3) between and we obtain the following result:

The light intensity corresponds to the spectrumof the light source. Since the

absorption

(or opticaldensity) equals the decimal logarithm of the inverse of thetransmittance (where transmittance is defined by

m λ( )

m λ( ) c

m λ( ) 10ln c ε λ( )⋅ ⋅=

ε λ( )

dφ λ x,( ) 10ln c ε λ( ) φ λ x,( )dx⋅ ⋅ ⋅–=

Fig. 1 Absorption in an infinitely thin layer.

Positive Direction

φ φ + dφ

x 0= x d=

φ λ d,( ) φ λ 0,( ) e10ln cd ε λ( )⋅ ⋅– φ λ 0,( ) 10

cd ε λ( )⋅–⋅= =

φ λ 0,( )A λ( )

), we have:

Beer pointed out

that what really counts for theabsorption is not the individual values of the concentration and of the path length , but rather the product whichrepresents a number of molecules per unit of area (or surfacedensity). Hence the transmittance can be expressed:

This equation holds as long as the medium is transparent,purely absorbing, non diffusing and non fluorescent.Furthermore, there must be no interaction between theabsorbing molecules.

Equation (6) expresses only the transmission factor in theabsorbing medium. The difference between the refractiveindices of the medium and the air causes multiple internalreflections (see Fig. 2) which modify the global transmittance

of the medium. From literature

we know that theintensity of each reflected ray follows a geometric sequence.The global transmittance is obtained by summing up thecontribution of each reflected ray and the result is given inequation (7) in terms of and the internal reflectionfactor .

According to Wyszecki,

is small so can beapproximated by:

φ λ d,( ) φ λ 0,( )⁄

A λ( ) cd ε λ( )⋅=

cd q cd=

T λ q,( ) φ λ q,( )φ λ 0,( )------------------- 10

q ε λ( )⋅–= =

Tg λ( )

T λ( )r

Fig. 2 Derivation of the transmittance of a transparent substrate(equation (7)).

Tg λ( ) 1 r–( ) 2T λ( )⋅

1 r–( ) 2T λ( ) r T λ( )⋅( ) 2⋅ ⋅

1 r–( ) 2T λ( ) r T λ( )⋅( ) 4 …+⋅ ⋅

+T λ( ) 1 r–( ) 2⋅

1 r T λ( )⋅( ) 2–

------------------------------------

Tg λ( ) T λ( ) 1 r–( ) 2⋅

1 r T λ( )⋅( ) 2–

------------------------------------=

r Tg λ( )

Tg λ( ) T= λ( ) 1 r–( ) 2⋅

3. The fluorescence phenomenon

In order to present in a few words the basic principles ofmolecular fluorescence,

let us consider a theoreticalmolecule having two electronic energy states, (groundstate) and (excited state). Each electronic state has severalvibrational states (see Fig. 3). Incident polychromatic light(photons) excites the molecules which are in state andmakes them populate temporarily the excited vibrationalstates of (Fig. 3 (a)).

A vibrational excited state has an average lifetime ofonly second. The molecule rapidly loses its vibrationalenergy and goes down to the electronic energy state . Thisrelaxation process is non radiative, and it is caused by thecollisions with solvent molecules to which the vibrationalenergy is transferred. This induces a slight increase of thetemperature of the medium. The excited state has alifetime varying between and second. Now, thereare two ways for the molecule to give up its excess energy.One of them is called

internal conversion

, a non radiativerelaxation whose mechanism is not fully understood. Thetransition occurs between and the upper vibrational stateof (Fig. 3 (b)), and the lost energy rises the temperature ofthe medium. The other possible relaxation process isfluorescence. It takes place by emitting a photon of energycorresponding to the transition between and a vibrationalstate of (Fig. 3 (c)). The remaining excess energy withrespect to is lost by vibrational relaxation. To quantify theamount of energy emitted by fluorescence, the

quantum yield

is introduced as the rate of absorbed energy which is releasedby radiative relaxation.

The wavelength band of absorbed radiation which isresponsible for the excitation of the molecules is called the

excitation spectrum. This spectrum consists of lines whosewavelengths correspond to the energy differences betweenexcited vibrational states of and the ground electronic

1015–

6–10

Fig. 3 Energy level diagram of (a) absorption, (b) non radiativerelaxation and (c) fluorescent emission. Note that for the resonanceline, absorbed and emitted photons have the same energy.

(a) Absorption (b) Non radiativerelaxation

(c) Fluorescence

ResonanceLine

state (according to the energy difference produced bythe absorption of a photon of wavelength : ,where is Planck’s constant and is the speed of light). Thefluorescence emission spectrum (or fluorescence spectrum)on its part, consists of lines which correspond to the energydifferences between the electronic level and thevibrational states of . The multitude of lines in both spectrais difficult to resolve and makes them look like continuousspectra. Note that the fluorescence spectrum is made up oflines of lower energy than the absorption spectrum. Thiswavelength shift between the absorption band and thefluorescence band is called the Stokes shift. There is aparticular case in which the absorbed photon has the sameenergy as the one re-emitted by fluorescence; it is called theresonance line.

The shape of the fluorescence emission spectrum doesnot depend on the spectrum of the exciting light, but on theprobability of the transition between the excited state andthe vibrational states of . Often, the fluorescence spectrumlooks like a mirror image of the excitation spectrum;7 this isdue to the fact that the differences between vibrational statesare about the same in ground and excited states.

Experience shows that fluorescence is favoured in rigidmolecules which contain aromatic rings.11 This can easily beunderstood since a rigid molecule has less possibilities ofrelaxing by a non radiative process. In fact, the lower theprobability of non radiative relaxation is, the higher thequantum yield. Hence, a rise in the medium’s viscosityinduces a higher fluorescence. In the particular case of inks,the liquid substance fluoresces less than the printed onewhose molecules have less degrees of freedom. On the otherhand, a rise of the ambient temperature implies a higherprobability of non radiative relaxation due to collisions withother molecules and a drop in fluorescence is observed.

The fluorescence spectrum is measured with afluorescence spectrometer.6 A sample of the unknownfluorescent substance is excited with a monochromatic lightbeam whose wavelength is within the excitation band of themolecule. The emitted light is analysed and the resultingspectrum is the fluorescence spectrum. Its amplitude ismaximal when the wavelength of the incident lightcorresponds to the maximum of absorption of the fluorescentmolecule. We denote by the normalized fluorescencespectrum whose integral equals 1.

The method we used to determine the quantum yield isdescribed in the literature.8 It is based on a measurementmade relatively to a standard fluorescent substance of knownquantum yield. To be reliable, the spectral emitting propertiesof the chosen standard fluorescent substance must closelymatch those of the unknown fluorescent substance.

E0 E∆λ E∆ hc( ) λ⁄=

f λ( )

The quantum yield of the unknown substance is givenby:8

In this equation the subscript u stands for unknown andthe subscript s for standard. is the absorption at theexcitation wavelength and is the quantum yield. Therefractive indices of the solvent of the standard fluorescentsubstance ( ) and of the medium of the unknownfluorescent substance ( ) are also taken into account. Thevariable is the total amount of energy emitted byfluorescence. This value is computed by integrating thespectrum emitted by fluorescence during the experiment.

At high concentrations the behaviour of the fluorescentsubstance is no longer linear. The absorption is too large andno light can pass through to cause excitation. Hence, to makea reliable quantum yield measurement the absorption has tobe smaller than 0.1 over the whole spectrum ( ).Temperature, dissolved oxygen and impurities reduce thequantum yield, too, and therefore they also reduce thefluorescence; this phenomenon is called quenching. In ourstudy we will suppose that no quenching occurs.

4. Infinitely thin layer

In order to establish a mathematical formula to predict thebehaviour of a transparent medium which containsfluorescent molecules, we consider a slice of thickness .We denote by the natural absorbance index of thefluorescent molecules and by their quantum yield in thismedium. In this model, only the positive direction ofpropagation is taken into account, hence the name “onechannel” model.

The intensity variation of the light emerging in thepositive direction has two components. The first, , is

AsFun2

AuFsn02

------------------ Qs⋅=

A λ( ) 0.1≤

dxm λ( )

Fig. 4 Absorption and emission in an infinitely thin fluorescentlayer

Positive Direction

φ φ + dφ

dφdφ1 λ( )

due to the light which has been absorbed according to Beer’slaw: (see equation (1)). Thesecond component, , is the light emitted byfluorescence. The fluorescent molecules emit a fraction ofthe energy absorbed in the excitation spectrum and spreadit over the whole emission band defined by the normalizedfluorescence spectrum . Due to the fact that fluorescentemission is made in all directions of space, only one half ofthe energy goes into the positive direction. Hence, thequantum yield must be divided by two. The secondcomponent is therefore given by:

The integral between square brackets multiplied by equals the amount of absorbed energy. Equation (10) leads tothe following differential form which is an extension ofBeer’s law (equation (1)):

This can be simplified due to the fact that we work witha finite number of wavelength bands whose widths are , sothat the integral is replaced by a finite sum. The new relationis given in equation (12) where the index runs through thewavelength bands.

Writing equation (12) for each of the bands leads to asystem of linear differential equations with constantcoefficients which can be put into a matrix form. If we denote

we obtain equation (13).

dφ1 λ x,( ) m λ( ) φ λ x,( ) dx–=dφ2 λ x,( )

f λ( )

dφ2 λ x,( )

dφ2 λ x,( ) Q2---- f λ( ) m µ( ) φ µ x,( ) µd

∆∫ dx⋅=

dφ λ x,( ) m λ( ) φ λ x,( ) dx–

Q2---- f λ( ) m µ( ) φ µ x,( ) µd

∆∫ dx⋅+

dφ λi q,( ) m– λ i( ) φ λi x,( ) dx

Q2---- f λ i( ) m λ j( ) φ λj x,( ) λ∆

j ∆∈∑⋅+ dx

Ki j, m λ j( ) Q2---- f λ i( ) λ∆=

The fact that an emitted photon has less energy than theabsorbed one implies that for ; hence thematrix is triangular.

The solution of equations such as (13) has already beeninvestigated by mathematicians.1 Systems of differentialequations whose general expression is (where

is the constant square matrix of equation (13) and is thecolumn vector containing , ... , ) admit assolution, when is integrated between and :

The vector is the spectrum of the incident light(light source) and is the spectrum of the light emergingfrom a slice of thickness of the fluorescent medium. Theexponential of the matrix is defined as follows:

Note that the natural absorption index isproportional to the concentration of the fluorescentsubstance (see equation (2)). In section 2 the surface density

(product of concentration and path length) was introduced.By factorising out in equation (13), we can make thechange of variable and obtain:

Ki j, 0= λ i λ j≥

xddΦ M Φ⋅=

M Φφ λ1 x,( ) φ λn x,( )

Φ d( ) Md( )exp Φ 0( )⋅=

Φ 0( )Φ d( )

Md( )exp Md( ) i

i!-----------------

m λ( )c

q– c– d=

M '10ln

-----------

ε λ1( ) 0 . . . . . 0

K'2 1, . .

. . . .

K'i 1, . K'i j, . ε λ i( ) . .

. . . .

. . . ⋅ .

K'n 1, . K'n j, . . . K'n n 1–, ε λn( )

dφ λ1 x,( )dx

-------------------------

dφ λi x,( )dx

------------------------

dφ λn x,( )dx

-------------------------

m– λ1( ) 0 . . . . . . 0

K2 1, . .

. . . .

Ki 1, . Ki j, . m– λ i( ) . .

. . . . .

. . . 0

Kn 1, . Kn j, . . . . Kn n 1–, m– λn( )

φ λ1 x,( )

φ λj x,( )

φ λn x,( )

where:

The solution (14) then becomes:

where:

and . (19)

We will call the fluorescence density matrix. Thespectrum resulting from the combined action of fluorescenceand absorption can be computed for each wavelength usingthe expression where and arerespectively components of and .

The solution given by equation (18) is a generalization ofBeer’s law: for a purely absorbing substance when nofluorescence is present, the matrix consists of the terms

on the diagonal and of zeros anywhere else. Thisallows us to simplify equation (18) and leads to equation (4).

Like in the case of Beer’s law, let us now have a look atthe influence of internal reflection. In this short analysis wedenote and is the internal reflectionfactor. According to Fig. 2 we can write the followingrelation between the spectrum of the emerging light and the spectrum of the source :

In media made of classical plastic compounds theinternal reflection factor is about 0.04, hence .This value is small enough to be neglected, so equation (20)can be approximated by:

This approximation is similar to the one used for the caseof Beer’s law (see equation (8)). Furthermore, by measuringdirectly (the spectrum of the lightemerging from the unprinted transparency) a simpler relationcan be used:

K'i j, ε– λ j( ) Q2---- f λ i( ) λ∆=

Φ q( ) qM– '( )exp Φ 0( )⋅=

M ' Mc-----–= qM– '( )exp qM– '( ) i

i!---------------------

λφ λ q,( ) φ λ 0,( )⁄ φ λ q,( ) φ λ 0,( )

Φ q( ) Φ 0( )

M '10ln ε λ i( )⋅

E qM '–( )exp= r

Φg q( )Φ 0( )

Φg q( ) 1 r–( ) 2E

I rE( ) 2 … rE( ) 2n …+ + + +( ) Φ 0( )

⋅ ⋅

r r 2 0.0016=

Φg q( ) 1 r–( ) 2E Φ 0( )⋅ ⋅=

Φ' 0( ) 1 r–( ) 2Φ 0( )=

Φg q( ) E Φ' 0( )⋅ qM '–( )exp Φ' 0( )⋅= =

5. Measuring the elements of the fluorescence density matrix

To compute the fluorescence density matrix fourelements have to be determined: the excitation spectrum, thedecadic extinction coefficient , the normalizedfluorescence function and the quantum yield . (Notethat contains discrete values of the functions and

Since the dye concentration in an ink is unknown, it isimpossible to determine the decadic extinction coefficient

. However according to equation (5), the absorptionspectrum and the decadic extinction coefficient are proportional and the proportionality factor is the dyesurface density . Note that each non-zero element of thefluorescence density matrix contains the factor (seeequations (16) and (17)). Since in equation (18) ismultiplied by , each occurrence of is multiplied by ,and this product equals the absorption . Hence, for agiven sample of absorption we do not need the actualvalues of and of , and we can work relatively to theabsorbance of a reference sample so that:

where is the proportionality factor between and.

The excitation spectrum is determined in a two-stepprocedure. In order to avoid deviations due to self absorption,the fluorescence measurement must be performed on asample whose maximal absorption is smaller than 0.1 overthe whole spectrum ( ). This means that lightemitted by fluorescence is not reabsorbed by anothermolecule of the sample. At first, the whole absorptionspectrum of our sample is measured with aspectrophotometer. This instrument uses a monochromaticcollimated light beam which goes through the transparentsample before reaching a light detector. Since only a smallfraction of the fluoresced light passes through the entranceslit of the detector, the deviation induced by the fluorescentemission can be neglected.

In the second step the location of the excitation spectrumwithin the absorption spectrum is determined. This can bedone, once we have an a priori knowledge of the approximateposition of the fluorescence spectrum (for example, by apreliminary measurement), using a fluorescencespectrometer. This device has two monochromators: the firstone is used to generate a monochromatic light beam whichexcites the sample, and the second monochromator is used toanalyse the light emitted by the sample. In our presentmeasurement, the second monochromator is set to a fixedwavelength which is supposed to be within the fluorescence

ε λ( )f λ( ) Q

M ' ε λ( )f λ( )

ε λ( )A λ( ) ε λ( )

qM ' ε λ i( )

M 'q ε λ i( ) q

A λ i( )A λ i( )

q ε λ i( )A' λ i( )

A λ i( ) qε λ( ) q'A' λ i( )= =

q' A' λ i( )A λ i( )

A λ( ) 0.1≤

A λ( )

spectrum (the a priori knowledge). The first monochromatorsweeps the whole spectrum and the intensity of the emittedlight is recorded. This provides the excitation spectrum andits location.

To determine the normalized fluorescence function afluorescence spectrometer is needed. The sample is excitedwith a monochromatic light beam whose wavelengthcorresponds to the maximum absorption in the excitationspectrum. Since the shape of the fluorescence emissionspectrum does not depend on the excitation wavelength (seesection 3), the normalized fluorescence function is easy tocompute by dividing the measured fluorescence spectrum byits integral value which corresponds to the fluoresced energy.

Once the excitation spectrum and the fluorescencefunction have been measured, a standard fluorescentsubstance can be chosen in order to calculate the quantumyield using a relative measurement method8 (see section 3).In order to perform a comparison, the location of theexcitation spectrum and the location of the fluorescencespectrum of the standard substance must correspond to thoseof our sample. Based on these criteria, the standard substanceis chosen from tables given in literature.4

Within the excitation spectrum we must select a singlewavelength which gives the highest possible fluorescence inboth the standard substance and our sample. These twosubstances are excited using the same fluorescencespectrometer at the selected wavelength and the spectrum ofthe fluoresced light is measured. By integrating thefluorescence spectra of the standard substance and of ourunknown sample, we get the respective energy amounts and emitted by fluorescence. Since the excitation spectraof both substances are known, we have the respectiveabsorption factors and at the selected excitationwavelength. Finally, the quantum yield of our sample iscalculated using equation (9). Note that three values must befound in the literature: the quantum yield of the standardsubstance,8 the refraction index of the medium containingit and the refraction index of our sample’s medium.

This experimental determination of the quantum yield israther difficult to perform. We preferred therefore in somecases to estimate the quantum yield by using a best-fit methodapplied on a test sample. Note that this choice reduces thenumber of experiments to be performed but does no longerguanrantee that the real physical quantum yield is used.

6. Results of the spectral and colorimetric pre-dictions

Three different inks were used at various concentrations andon three different supports. The spectra of all samples weremeasured and compared to the results given by the prediction

As AuQu

model. The three fluorescent inks are the magenta ink used inHP 500 C desktop printers, and the orange and yellow inksused in fluorescent markers. Transparencies from threedifferent manufacturers (3M, Epson and Sihl) were used assupport. We measured all the elements necessary to establishthe fluorescence density matrix for these inks: the absorptionspectrum, the fluorescence spectrum, the absorption band andthe emission band. The quantum yield was either measured orestimated by applying a best-fit procedure. Note that thequantum yield of a given ink depends on the support.

The reference absorption spectra of the inks weremeasured on samples obtained by applying the ink uniformlyon the transparent substrate. These measurements wereperformed on a Variant spectrophotometer.

We illustrate our method with the results obtained for themagenta ink. The absorption spectrum of the magenta ink isgiven in Fig. 5.

The excitation band of the magenta ink is 450-600 nm,and its emission band is 570-800 nm. In order to measure thequantum yield of the magenta ink, Rhodamine 6G dissolved

A' λ( )

450 500 550 600 650 700nm0

Fig. 5 Absorption spectrum of the magenta ink.A' λ( )

Fig. 6 Fluorescence emission spectrum of magenta ink excited at550 nm.

550 600 650 700 750 800nm

Emission spectrum of magenta ink

Counts

in ethanol was chosen as the reference substance. Themeasurement was performed on a fluorescence spectrometerPerkin-Elmer LS50B as described in section 5. The quantumyield for the 3M transparency was computed with equation(9) and we obtained . The quantum yield for theother substrates were estimated by a best-fit algorithm. Thesedata are summerized in the first column of Table 1. Themeasured fluorescence spectrum of the magenta ink is shownin Fig. 6. With these elements the fluorescence density matrix

is computed using wavelength bands of widthnm.

Our new model was applied to predict the spectra ofuniform patches having different concentrations (fourdifferent patches for each ink-substrate combination). Theproportionality factor (see equation (23)) was computedrelatively to reference samples by comparing the absorptionof the reference sample and the absorption of thecorresponding unknown sample in a wavelength band whereno emission takes place. The spectra of the unknown sampleswere measured by transparence using an Oriel 77400radiospectrometer combined with an Oriel integrating sphere.The light source is the measurement spot of the Spotlightlight table from Light Source. Its relative spectrum has been

Magenta Yellow Orange

Absorption band (nm) 450-600 310-520 310-580

Emission band (nm) 570-800 480-700 550-750

Quantum yield on transparency:

3M 0.12* 0.15 0.6

Epson 0.5 0.8 0.5

Sihl 0.3 0.6 0.6

Table 1: Absorption band, emission band and quantum yields ofthe inks. The only measured case is marked by *; the other caseswere estimated by a best-fit algorithm.

Q 0.122=

M 'λ∆ 10=

Fig. 7 Relative spectrum of the light source.450 500 550 600 650 700

Counts

measured and is shown in Fig. 7. The matrices were computed numerically (see section 4). Fig. 8 illustratesthe obtained results: the continuous line is the measuredspectrum, the dotted line shows the prediction whenfluorescence is taken into account and the dashed linecorresponds to the prediction given by Beer’s law. Note thatthe three lines almost perfectly superpose when onlyabsorption takes place. In the emission band, only smalldeviations are observed between the measured spectrum andthe spectrum predicted when fluorescence is taken intoaccount. The colour deviation between predicted andmeasured spectrum is CIE-Lab whereas thecolour deviation between the measured spectrum and thespectrum computed with Beer’s law is .

The colorimetric deviations of all samples have beencomputed in the CIE-Lab colour space, and the results aresummarized in Table 2. In order to highlight the contributionof fluorescence, we give the average deviation of the foursamples for each ink-substrate combination when using ourmodel (second column), and when only Beer’s law is takeninto account (third column). As it can be seen in the table,prediction improvements are significant only when thequantum yield is relatively high. Note also that the quantumyield depends not only on the inks but also on the substrate:the quantum yields of the magenta and yellow inks aresignificantly reduced when these inks are applied on a 3Mtransparency. This is also confirmed by visual observation:3M transparencies reduce the observed colourfulness formagenta and yellow inks, but do not influence thecolourfulness of the orange ink.

7. Conclusions

Our proposed new spectral prediction method is based on amathematical formalism which generalizes Beer’s law. Thismodel requires measuring the transmittance spectra, the

q– M '( )exp

∆E 1.92=

∆E 8.59=

Fig. 8 Measured spectrum of the magenta ink on Epsontransparency (continuous line); spectrum predicted with ourmethod taking fluorescence into account (dotted line); andspectrum predicted by Beer’s law (dashed line).

450 500 550 600 650 700nm0

normalized fluorescence spectra and the quantum yield of thefluorescent inks. In contrast to previous methods, with ourapproach, once the fluorescence density matrix is computed,prediction can be made for different illuminants and differentink concentrations.

Using this method we predicted the transmittance spectraof uniform samples. The average prediction error is about

with a maximum of . Since inkfluorescence is also taken into account, the predictionaccuracy is improved by about in comparison withBeer’s law.

This model does not take the whole complexity of thefluorescence emission phenomenon into consideration. Forinstance, it does not include the quenching effects due to highconcentrations. However, it enables the spectra of fluorescentinks to be predicted qualitatively and quantitatively underdifferent illumination and concentration conditions.

Type of sample

Average using

the new model with fluorescence

Average using

the model without

fluorescence

Quantum yield

Magenta on 3M

1.22 1.54 0.12

Magenta on Epson

1.83 7.08 0.5

Magenta on Sihl

1.56 4.68 0.3

Yellow on 3M

1.29 2.39 0.15

Yellow on Epson

1.29 6.94 0.8

Yellow on Sihl

1.52 6.24 0.6

Orange on 3M

2.13 6.81 0.6

Orange on Epson

1.44 5.16 0.5

Orange on Sihl

1.84 6.70 0.6

Table 2: Colour deviations in CIE-Lab between predictedand measured spectra for inks at different densities.

∆E 1.56= ∆EMax 3.08=

∆E 4=

∆E ∆E

8. Acknowledgment

We would like to thank Dr. J.E. Moser of the Laboratoire dePhotonique et Interfaces (LPI) of the EPFL for his help anduseful advice, and for the interesting discussions inphotochemistry. We also like to thank the Swiss NationalFund (grant No. 21-040541) for supporting the project.

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A ‘one channel’spectral colour prediction model for transparent fluorescent inks on a...

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