Lesson 03 Radiometry

8/12/2019 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



2

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



3

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



5

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



7

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



10

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



11

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



12

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,



14

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



17

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



19

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



20

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



22

Blackbody RadiationBlackbody Radiation

Figure 2.29 Spectral photon radiance vs. wavelength for blackbody

temperatures from 1200K to 6000K



23

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



25

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

Lesson 03 Radiometry

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