1. The document discusses how light interacts with colored paint films. When light hits the paint surface, some of it is reflected at the surface, while the rest enters the paint layer and undergoes absorption and scattering. 2. Absorption and scattering of light in the paint layer depends on factors like the refractive index of the paint resin, pigment particle size, and the difference in refractive index between pigment particles and the resin. 3. Both surface reflection and scattering contribute to the overall appearance and color of the painted surface. The balance between these factors can be described using polarization or gonio-photometric reflection curves.

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2. Introduction
• Consider a beam of white light incident on the surface
• of a coloured paint film.
• As soon as the light meets the paint surface
 the beam undergoes refraction,
 and some of the light is reflected.
• The refracted beam entering the paint layer then undergoes
 absorption
 and scattering,
 and it is the combination of these two processes
 which gives rise to the underlying colour of the paint layer.
• In order to have some appreciation of the optical factors which give the
surface overall appearance (including colour and gloss or texture)
• we need to outline the laws
 that affect the interactions of the light beam with the surface.
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3. Introduction
the white light beam, considered as a bundle of waves with
wavelengths covering
the range 400–700 nm,
can also be considered as a wave-bundle in which the
waves have components
which vibrate in planes mutually
at right angles to one another
along the line of transmission.
If the wave vibrations are confined/restricted /bound/held to
one plane we describe the radiation as being
plane polarised.
Polarisation effects are important when we consider
reflections from glossy surfaces and mirrors.
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4. 4
Refraction of light
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5. Snell’s law
Refraction into the interior of the film takes place according to
Snell’s law,
Which states that
when light travelling
through a medium of refractive index n1 encounters
and enters a medium of refractive index n2
then the light beam is bent
through an angle
according to Eqn 1.11:
where i is the angle of incidence and
r is the angle of refraction
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Refraction of light
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• A typical paint resin
has a refractive index
similar to that of
ordinary glass (n =
1.5)
• and so a beam of
radiation incident on
the surface at 45°
• will be bent towards
the normal by 17°
• to a refraction angle
of approximately 28°.

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Refraction of light
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• The refraction angle depends
on
 the wavelength;
• the ability of glass to refract
blue radiation more than red
radiation is apparent in the
production of a visible
spectrum when
 white light is passed through a
glass prism.
• Refractive indices are therefore
normally measured using
 radiation of a standard
wavelength
 – in practice, sodium D line
radiation (yellow-orange light
of wavelength 589.3 nm).

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Surface reflection of light
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Fresnel’s law
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• A light beam incident
normally (vertically) on a
surface or any boundary
• between two phases of
differing refractive index
will suffer partial back-
reflection according to
• Fresnel’s law (Eqn 1.12):
where r is the reflection factor for un-
polarised light and n is n2/n1.

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Fresnel’s law
• If the incident light beam is white then
 the light reflected from the surface will also be white
 (white light needs to undergo selective absorption before
it appears coloured).
• This small percentage of white light reflected from
the surface affects
 the visually perceived colour,
• and instrumentally measured reflectance values
should indicate whether the specular reflection is
 included (SPIN) or
 excluded (SPEX).

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Surface reflection of light
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• For the air (n = 1) and
• resin layer (n = 1.5)
interface the total
surface reflection at
▫ normal angles is about
4% (r = 0.04).
• At angles away from
the normal, however,
this
 surface or specular
(mirror-like) reflection
varies
• depending on
 the polarisation of the
beam relative to the
surface plane (Figure
1.23).

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Surface reflection of light
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• The curves in this diagram
show that
• the reflection of the
perpendicularly polarised
component becomes
 zero at a certain angle (the Brewster
angle),
• and the reflected light at this
angle is polarised
 in the one direction.
• The reflection of both
polarised components
becomes
 equal at normal incidence (0°),
• and again at the grazing angle
(90°), at which point the
surface reflects
 virtually 100% of the incident light
(surfaces always look glossy at high
or grazing angles).

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Surface reflection of light
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• Thus light reflected
from most surfaces is
 partially polarised.
• This is why Polaroid
glasses are useful for
 cutting out glare from
wet roads
 when driving,
 and for seeing under
 the surface of water on
a bright day.

14. Light scattering and diffuse
14
reflection
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Light scattering and diffuse
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reflection
• Part of the light beam is not specularly reflected at
the surface but
 undergoes refraction into the paint layer.
• This light will encounter pigment particles, which
will
 scatter it in all directions.
• The extent of this scattering will depend on
 the particle size
 and on the refractive index difference between the
pigment particles
 and the medium in which
 they are dispersed, again according to Fresnel’s laws.

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Light scattering and diffuse
reflection
• With white pigments like
• titanium dioxide (n > 2) the scattering will be
 independent of wavelength,
 and most of the incident light will be scattered in random directions.
• A high proportion will reappear at the surface and give rise to the
diffuse reflected component;
 with a good matt white the diffuse reflection can approach 90% of the
incident light.
• White textile fibres and fabrics produce a high proportion of
diffusely reflected light, either because of the scattering at the
 numerous interfaces in the microfibrillar structure of natural fibres like
cotton, wool and silk or,
• in the case of synthetic fibres, from the presence of titanium
 dioxide pigment in the fibres.

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polar reflection or gonio-photo-metric
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reflection curve
In practice there will be a balance between
specular and diffuse reflected light
• which can be described by the
polar reflection or gonio-photo-metric reflection curve
• Shown in Figure 1.24).

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TO Assess the Gloss and Coloristic
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Properties
• To assess the gloss, determined
by the proportion of the
 Specular component,
• the sample should be viewed at
 an angle equal to the incident, i.e. at
60°
• for the case illustrated in Figure
1.25.
• The extent of the diffuse
component
• (and any colour contribution)
is then assessed by
 viewing at right angles to the surface
(that is,
 at an incident angle of 0°,
• Figure 1.26).

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Light scattering and diffuse
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reflection
• Thus the direction of reflected light plays a large part in
the
 appearance of a surface coating.
• If it is concentrated within a narrow region at an angle
equal to the angle of incidence
 the surface will appear glossy, i.e. it will have a high specular reflection.
• Conversely if it is reflected indiscriminately/ random
/jumbled / multifarious at all angles it will have
 a high diffuse reflection and will appear matt.
• Gloss is usually assessed instrumentally at high angles
• (60 or 85°)
• as the specular component is more important at such high angles
▫ (even a ‘matt’ paint surface shows some gloss at high or grazing angles).

20. Absorption of light 20
(Beer–Lambert law)
If the paint layer contains coloured pigment particles
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the light beam travelling through the medium will be
 partly absorbed and partly scattered(Figure 1.1).
Some particles are so small (< 0.2 mm) that they can be considered
to be effectively in solution, and their light-absorption properties
can be treated in the same way
as those of dye solutions which absorb but do not scatter light.

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Transmission of Light through dye
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solutions
• The transmission of
light of a single
wavelength
(monochromatic
radiation)
through dye solutions or
dispersions of very small
particles
• is governed by two laws:
1. Lambert’s or Bouguer’s
law (1760),
1. Beer’s law (1832),

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Lambert’s or Bouguer’s law (1760)
• which states that layers of equal thickness
of
the same substance
 transmit the same fraction of the incident
monochromatic radiation,
whatever its intensity

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Beer’s law (1832)
• which states that the absorption of light is
proportional to the
number of absorbing entities (molecules) in its
path;
• that is, for a given path length,
the proportion of light transmitted decreases
with the concentration of the light-absorbing solute.

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The Beer-Lambert law
• A = a(λ) * b * c
 where A is the measured
absorbance,
 a(λ) is a wavelength-dependent
absorptivity coefficient,
 b is the path length,
 and c is the analyte concentration.
• When working in concentration
units of molarity, the Beer-
Lambert law is written as:
A= ε *b*c
where ε is the wavelength-
dependent molar absorptivity
coefficient with units of M-1 cm-1.

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The Beer-Lambert law
• The Beer-Lambert law can be
derived from an approximation for
 the absorption coefficient
 for a molecule
 by approximating the molecule
 by an opaque disk
• whose cross-sectional area,σ ,
represents the effective area seen by
 a photon of frequency w.
• If the frequency of the light is far
from resonance,
 the area is approximately 0,
• and if w is close to resonance
 the area is a maximum.
• Taking an infinitesimal slab, dz, of
sample

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The Beer-Lambert law
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 Io is the intensity entering the
sample
 at z=0,
 Iz is the intensity entering the
infinitesimal slab at z,
 dI is the intensity absorbed in
the slab,
 and I is the intensity of light
leaving the sample.
 Then, the total opaque area on
the slab due to the absorbers
is σ * N * A * dz. Then, the Integrating this equation from z = 0
fraction of photons absorbed to z = b gives:
will be σ* N * A * dz / A so,

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The Beer-Lambert law
• Since N (molecules/cm3) * (1 mole /
6.023x1023 molecules) * 1000 cm3 / liter = c
(moles/liter)
• and 2.303 * log(x) = ln(x), then

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The Beer-Lambert law
• Suppose that we were to
measure the absorption of
green light by a purple
dye solution
• contained in a
spectrophotometer cell
(cuvette) of total path
length 1 cm,
• And that the solution
absorbed 50% of the
incident radiation over the
first 0.2 cm;
• then the
• light transmittance through
the cell would vary as
shown in Table 1.5.

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The Beer-Lambert law
• Each 0.2 cm layer of
solution decreases
the light intensity
• by 50%,
• as required by the
Lambert– Bouguer law.
• The quantity log
(1/T), known as the
absorbance,
• increases linearly with
• thickness or path
length,
• whilst the intensity
decreases exponentially
(Figure 1.27).

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The Beer-Lambert law
• A plot of Beer’s law behaviour at fixed path
length would show a similar linear dependence
• of absorbance A with concentration. In fact the
combined Beer–Lambert
• law is often written as Eqn 1.18:

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The Beer-Lambert law
• where the proportionality constant ε is known as
the absorptivity; if the concentration
• is in units of moles per unit volume (litre), it is
known as the molar absorptivity.
• The combined Beer–Lambert law can
alternatively be written as Eqn 1.19:

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The Beer-Lambert law
• Measurements of absorbance are widely used,
through
 the application of the Beer–Lambert law,
 for determining the amount of coloured materials in
solution,
 including measurements of the strengths of dyes .
• In practice deviations from these laws can arise from
both
▫ instrumental
▫ and solution (chemical) factors,
• but discussion
 of these deviations is outside the scope of the present
treatment