Lesson 03 Radiometry
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Radiometry is the science and technology of themeasurement of electro-magnetic radiant energy.
Fundamental question: How much optical
energy from a source is collected on a detector
surface for a particular configuration?
This requires the understanding of:
1. Solid Angle
2. Radiant Flux Transfer
3. Point Sources4. Extended Sources
5. Blackbody Sources
6. Units
7. Emissivity
Radiometry of incoherent thermal sources small
angle approximately almost always.
RadiometryRadiometry
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Nomenclature for Radiometric TermsNomenclature for Radiometric Terms
e
Sometimes a p is used as the subscript for photon quanties.
No photometric units; i.e. lumens, candle, etc.
Quantity
radiant energyradiant power (Flux)
radiant intensity
irradiance (Flux density)
radiant Exitance
radiance
photon energy photon flux
photon flux intensity
Incident Photon
flux density (irradiance)
photon flux exitance
photon flux sterance (radiance)
Qe
Ie
Ee
Me
Le
Q q
Iq
Eq
Mq
Lq
joule
Watt
Watt/steradian
Watt/cm2
Watt/cm2
Watt/cm2-sr
photon or Quantum
photon/sec
photon/sec - sr
photon/sec - cm2
photon/sec - cm2
photon/sec - cm2
Symbol Units
q
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Solid AngleSolid Angle -- DiagramDiagram
0
2
2 2
2π
0
0 0
dA ρdθρ sin θd
dA ρ sin θdθddω
ρ ρ
sin θdθ 2π 1 cos θ
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Radiant PowerRadiant Power
Radiant Power Collected:
Exit Pupil - AexEntrance Pupil -Aen
Le
Adobject
R
e en L A
Detector radiant
power or
irradiance here
chiefray
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1
2
Aen
sensor
source
As
LR
Radiant Power TransferRadiant Power Transfer
1 2
2
Source radiance
Entrance pupil area, but need projected
area/solid angle so cosines
cos cos
en
s en
LA
L
A
A A L
R
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The relationship
between radiant
exitance (M) and
radiance (L) is in
general complex,
and depends on
the angular
distribution of
radiance, L(θ1,ø).
1 1 2
2
2 1 1
1 1 1 1
π2π
21 1 1 1
0 0
π2 2
1
0
L θ , A cos θ dAdR
dA Rd Rsinθ dθ
d L θ , Asin θ cosθ dθ d
d L θ , Asinθ cosθ dθ
Assume that L is independent of direction (this is called lambertian), then
sin θ2πL A
2
Substitut
ing M and solving:
A
The error most often committed in radiometry is in this converstion.Don't make the frequent mistake of forgetting the cosine and using 2π!
M πL
dA2
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Irradiance vs. Range/Source RadiusIrradiance vs. Range/Source Radius
0.0001
0.001
0.01
0.1
1
10
0.1 1 10 100
log(R/a)
l o g ( E / L * P i )
exact
Series2
1)a/R (
1LE
2
s
ee
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Irradiance vs.Irradiance vs.Range/Source RadiusRange/Source Radius
• Case I:
–Large distance ( R/a>>10)
» Irradiance fall off as 1/ R2 (inverse law)
• Case II:
–Large source area ( R/a<<1)
» Irradiance is L on receiver
2
2e
L a E
R
e E L
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boundarymediaanyinconstant:Prove2
n
L
u
en A pupilEntrance
Throughput of an Optical SystemThroughput of an Optical System
F.O.V.angular solid
pupilentranceof area
étendue constant
&
2
2
2
u
An
A
ynu yun
y, uu , y
opticsorder1st
Brightness is conserved
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Proof of ConstantProof of Constant
constantis :Proof 2n
L
n1 n2
dASource
radiance
I 2
I 1r
L2
L1
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
1 1 1
2 2 21 2
since
sin ;
sinsin
s o ,sin
sin
sin;
d I dI d
d I dI d d I dI d
d I dI d
d I dI
d I dI d d
For source L1, solid angle from dA produces is
Similarly for source L2:
In the sagittal plane,
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22
11
1
2
222111
2211
cos
cos
coscos
sinsin
dI I
dI I
n
n
dI I ndI I n
I n I n
2
1 1 1
22 2 2
cos
cos
LA
d L dA I d
d L dA I d
From Snell’s law
Radiative transfer of power ( ) or photons:
but power is transferred or equal if transmission is
accounted, or power conserved
22
1112
cos
cos
d I dA
d I dA L L
substitute for 21 / d d
2
1
1
2
2
2
1
2
1
2
12
Snellof aldifferenti
22
11
Snell
2
112
222
11112
cos
cos
sin
sin
sincos
sincos
n
L
n
L
n
n
n
n
L L
dI I
dI I
I
I L L
dI I I
dI I I L L
Inside a detector, radiance goes up by n2
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A s
source
R
Ad
detector
d
R/cos s
s
Cosine – to – the – fourth (cos4) law
L
R is along z - axis
42 cos s d
A A L
R
Exit pupil to detector
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The optics have assumed theradiance of the extended source.
Fills F.O.V. and aperture.
2 2 2
2
2
2
; but
;
sin
d s s o s
e e
s d o
e e
d s se oe e e
d
A A A L
R f R
A A L
f
A E L L
A f
Sources
detector
Ad
Finite
Ao=Aen
Extended
Extended SourcesExtended Sources
f
s
e
L
s
• Range (R) is not important if extended
• Conceptually, if source can be thought of as being made up of a
[i i] point source, and fills the F.O.V., the irradiance for each
point source fall as but the area of the source increases by
R2 ; so the power on detector is a constant or range
cancels out of equation of irradiance on detector.
2
1
R2 R A s
2
0A radiancesourcethetoequalisdetectoron
efl E
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Extended sources + optical elements are sources in IR.
Ad
Signal irradiance on detector is:
What is irradiance on detector from source?
Irradiance on
detector surface :
also:
A Ex
R
Entu
R
Detector
s
e L
1 2 3
1 2 3
o o o
o o o
Ex En
u
d
e E
u
2
1 2 3
2
sin
Atmospheric transmission
nuy nuy i
sin Ex
s s
e e
a
o o o o
Ex s Ene e o
Ex
Ex s Ene e a o
E L a
A
A E E
A
A E L u
A
Signal irradiance @ entrance pupil :
Signal irradiance @ exit pupil :
Lagrange InvariantЖ :
u
s constant
2 2sin sin En d u A u
2
2
sin
sin
d Ex s
e e a o
Ex
Ex
e Ex d s
e e a o
d
A u E L
A
E A E L u
A
2,
2
2
sin
1sin
4 /# 1
d s s
e e a o E L u
u F
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Extended source contribution from optics emission
Irradiance on detector from optical elements
Contribution of optics on detector to signal typically you
want this less than 10%.
Transmission of atmosphere is function of range!
, 2 2sin 1d o oe e E L u
The ratio of optical system irradiance to signal irradiance is:
2,
, 3
[1 ]od oee
d s se oe a
L E
E L
( )a fn R
• Transmission are all equal,
• Emissivity of glass is same -
Irradiance on detector from optics(background irradiance)
(3) (2), 2 23 2 3
(1) 21 2 3
sin sin
sin
d oe e eo o o
eo o o
E L u L u
L u
(1) (2) (3)Assume are all at the same temperatureo
e e e e L L L L
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Heated objects emit radiation - as a function (temp).
So far, we considered the object as an emitter having some
radiance - L
qe or
We need the concept of spectral source characteristics
Need to define source as a continuous function of
wavelength or frequency of light.
hc
or h
Denoted as spectral radiance
m sr cm
photon
m sr cm
Watt
T Lor T L qe
22 sec
),(),(
So the radiance is an integration over wavelength
( , )e e L T d
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Blackbody RadiationBlackbody Radiation
Figure 2.29 Spectral photon radiance vs. wavelength for blackbody
temperatures from 1200K to 6000K
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Blackbody Radiation (conBlackbody Radiation (con’’t)t)
Figure 2.31 Spectral radiance vs. wavelength for blackbody temperatures
from 1200K to 6000K
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Total radiance for a blackbody at temperature T , is the
integral over all ( ) wavelength.
42
12 )10(67.5ConstantBoltzmannStefan K cm
watt e
Called Stefan - Boltzmann Law
where
11 2 3
3
0
1.52 10 /
( ) ( , )
q
qq q
photons s cm K
T L T L T d
0
2
50
4 44
2 3
4
2
( ) ( , )
2
( 1)
2( )
15
( )
e e
x
Be
ee
L T L T d
c hd
e
k L T T
c hT watt
L T cm
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The decrease in the wavelength of peak exitance as
temperature increases is quantified in Wein’sdisplacement law.
The analytic relationship can be derived, at some wavelength,from the condition for the peak of the exitance function:
0
),(
T Le
This produces a constraint on the wavelength of
maximum exitance, :
( , ) 0q L T
maxT = 66 mK For example, a blackbody source at T – 300oK (roomtemperature) has its maximum exitance at approximately 9.7 μm.
If the source in question were at 1000oK, the value ofwould be at 2.9 μm.
It is interesting to note that the for the sun is near0.5 μm, very close to the peak of sensitivity of the human eye.
For photon
radiance:
WeinWein’’s Displacement Laws Displacement Law
)(
2898max
K T
K m
max
max
max
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Let λ short be the wavelength on the short λ side at which thePlanck curve obtains 50% of its maximum value.
Let λ long be the wavelength on the long λ side at which the
Planck curve obtains 50% of its maximum value.
The area under the curve from λ = 0 to λ short is 3% of
The area under the curve from λ short to λ long is 60% of The area under the curve from λ long to λ = ∞ is 37% of
The area under the curve from λ = 0 to λ max is 25% of
The area under the curve from λ max to λ long is 38% of
Other Useful Properties ofPlanck Functions (power)
4eσ T
π
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Change in radiance for a change in temperature
T
L
All optical radiation
34
3
( ) ( )
( ) 4
qee q
e e
T T L T or L T
L T T
T
e L ( λ ,T)
= 0
λ T λT = 2410 μm°K
∂ ∂
∂ ∂
Monochromatic BB radiation
For maximum contrast.
What is the maximum contrast with photon?
Temperature ContrastTemperature Contrast
2
/5
4
22.79
22.80
( , )( 1)
( , )( 1)
Be hc k T
q x
hc
c
L T e
L T e
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Emissivity - 0 < < 1, =1
• Absorptivity = emissivity
• Good absorber & emitter
Optical system looked at Earth, (400 km in height)
Earth
Sun
6000 K
2
1Sun angle
At what wavelength does Sun’s reflected irradiance (Ee) just equal
Earth’s exitance (Me)?
6000 K
300 K
Exitance
About the sunAbout the sun
300 K, 0.3 reflectivity
?c
BB
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Loss of Contrast (Ct) due to transmission is the ratio of contrast
with atmosphere influence to that with no atmosphere.
Target resolved,
AT > A IFOV - ground sampling distance radiance
through atmospheric, - transmission to sensor
L p = Path radiance between target
LB = Background radiance of sensor
LT = Target
Without atmosphere: T T
B B
L
L
Contrast with Atm
No Atm
( )
(100%)
T B
B
T B
B
LC L
L LC
L
T T
B B
L L L L
Contrast with Atm
Contrast w/o Atm
( )
(100%)T
C
C C
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P B
BT
P B
P B pT
L L L LC
L L
L L L LC
)()(
)(
Contrast Transmittance C ( ) /C (100%):
B
p P B
B
BT
B
P B
BT T
L
L L L
L
L L
L
L L
L LC
1
1
)(
Recall emissivity concept and relate to path and
background
Path emissivity
Background emissivity
Indicates blackbody emission
BB
P P p
BB
B B B
P
B
L L
L L
BB
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Assume no scattering; transmittance is one minus absorption
(good for IR) >3μm
Contrast transmittance is:
Iff the temperature of atmosphere is same as background
temperatures,
The background is black,
BB
P
BB
B L L
)1( B
The reduction in Contrast (CT) at sensor is equal toatmospheric transmittance in IR when there is
no scattering.
1 1 P
1
1(1 )
T BB p p
BB p B B
C L
L
11
11
1
pT
p p p
p
C
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Natural sourceNatural source
• Solar • reflectivity