ECE 425 CLASS NOTES – 2000 SECTION II – OPTICAL TOOLS
Introduction Radiometry Sources of Radiant Energy Photometry Radiometric Transfer
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Introduction
An imaging system consists of several subsystems
image light transmission display human scene acquisition vision source subsystem subsystem* subsystem subsystem
coder decoder optics neural network optics detector* electronics
retina* brain
• * points of signal transduction, optical <—> electronic From the light source to the image acquisition subsystem, we’re concerned with:
• how much energy gets through (radiometry)
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ECE 425 CLASS NOTES – 2000
• contrast and sharpness of the image (linear systems) Similarly from the display subsystem to the human vision subsystem
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ECE 425 CLASS NOTES – 2000
Radiometry
Radiometry involves the propagation of radiation in space and through optical apertures Need to use 3-D geometry to describe
3-D spherical coordinate system (r,θ,φ)
n θ Q
r P φ
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• 3-D vector with radius r
• angle to the surface normal θ (radians)
• azimuth angle φ (radians) Radiation propagates from a source into a conical volume
spherical cap area A
r source
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• Define a solid angle Ω (steradians)
2 Ω = Ar⁄ (steradians)
• Ω = 1 (unit solid angle) when the spherical area subtended by the cone = radius of the sphere Example: calculate the solid angles corresponding to a hemisphere and a sphere
• flat surface sources radiate into a hemisphere
• point sources radiate into a sphere
• set up integration over solid angle
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θ rsinθ dθ
r
P
• Assume the source P is an isotropic radiator • Element of solid angle
• circumference of element of solid angle (“belt”) around sphere is 2πr sinθ
• width of “belt” is rdθ 126
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• therefore area of “belt” is 2πr sinθ ⋅ rdθ and solid angle subtended by “belt” is
2πr sinθ ⋅ rdθ dΩ = ------2 r = 2πθsin dθ
• Total solid angle θ Ωθ()= ∫2πθsin dθ 0 = 2π()1 – cosθ
• For θ = π/2 (hemisphere)
Ω = 2π steradians
• For θ = π (sphere)
Ω = 4π steradians
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ECE 425 CLASS NOTES – 2000
Projected Area
• Useful in many radiometric calculations for both sources and detectors
• Area of surface element dA as viewed from an angle θ
n
θ dAcosθ θ
dA
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ECE 425 CLASS NOTES – 2000
Radiant Energy Quantities and Units
• Radiometric (valid for general case)
quantitiy symbol definition common units abbreviation radiant Q joule J energy Φ
radiant ∂Q joule per cubic w w = J m-3 density ∂V meter
∂Q radiant flux Φ = watt W ∂t
M ∂Φ M = (exitance) ∂A radiant flux watt per -2 density E square meter W m ∂Φ (irradi- E = ∂A ance)
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ECE 425 CLASS NOTES – 2000
quantitiy symbol definition common units abbreviation
radiant ∂Φ watt per I I = W sr-1 intensity ∂Ω steradian
2 ∂Φ watt per L = ∂Ω∂Acosθ radiance L steradian and W sr-1 m-2 ∂I = ∂Acosθ square meter
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• Photometric (special to visual sensing)
quantity symbol common units abbreviation lumen-second (tal- luminous energy Q lm s bot) lumen-second per luminous density w -3 cubic meter lm s m luminous flux Φ lumen lm M lumen per square (luminous meter exitance) (lux) -2 luminous flux den- lm m (lx) sity lm ft-2 (fc) E lumen per square foot (illuminance) (footcandle)
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quantity symbol common units abbreviation luminous intensity lumen per steradian I lm sr-1 (cd) (candlepower) (candela) candela per square meter (nit) nt luminance L candela per square π foot per steradian fL (footlambert)
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• Radiometric <—> photometric conversion
common quantity symbol definition units abbreviation
Φ ⁄ Φ -1 luminous KK = v e lumen lm W efficacy per watt luminous VVKK= ⁄ unitless — efficiency maximum
• Material properties
common quantity symbol definition abbreviation units ε ⁄ emissivity ε = MMblackbody unitless —
α αΦ⁄ Φ absorptance = a i unitless —
ρΦ⁄ Φ reflectance ρ = r i unitless —
τΦ⁄ Φ transmittance τ = t i unitless —
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Wavelength Notation
• Some quantities are differential with respect to λ, e.g. Lλ and Eλ
• Units of “per wavelength interval” • Must be integrated over λ to obtain total quantity within a given wavelength range
• Some quantities simply vary with λ, e.g. ρ(λ), V(λ) and τ(λ)
• Not integrated alone; used to weight another quantity
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Sources of Radiant Energy
Blackbody (BB)
• Perfect radiator and absorber
• Produces maximum M for any source at a given T
• Nonattainable, ideal source
• spectral radiant exitance M given by Planck’s Equation
2 2πhc M = ------(wavelength in meters) λBB 5 hc ⁄ ()λkT λ []e –1 (W-m-2-µm-1) C1 = ------⁄ ()λ - (wavelength in micrometers) 5 C2 T λ []e –1
where
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T is the blackbody’s temperature in Kelvin (K), h = 6.6256 x 10-34 (W-s2) Planck’s Constant k = 1.38054 x 10-23 (W-s-K-1) Boltzmann’s Constant c = 2.997925 x 108 (m-s-1) velocity of light λ = wavelength of radiation 8 -2 µ 4 C1 = 3.74151 x 10 W-m - m , and 4 µ C2 = 1.43879 x 10 m-K.
• Departure of a given source from a BB is its emittance • Measures the efficiency of a radiator or absorber ε ⁄ ≤≤ε λ =0Mλ MλBB , λ 1
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Sun
• Solar Irradiance at Top-Of-Atmosphere
2500
5900K BB at earth-sun distance )
-1 2000 MODTRAN m µ - -2 1500
1000
irradiance (W-m 500
0 0.4 0.8 1.2 1.6 2 2.4 wavelength (µm)
• Modeled well by a blackbody @ 5900K Solar energy propagates through the atmosphere to Earth’s surface
• Atmospheric transmittance creates spectral “windows” through which energy reaches the earth
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• Above 3µm wavelength, the Earth’s self-emitted radiation (BB at 300K) becomes significant
• Above 8µm, Earth’s self-emitted radiation dominates and solar radiation is insignifcant
wavelength radiation surface property name range source of interest Visible (V) 0.4 – 0.7µm solar reflectance Near InfraRed (NIR) 0.7 – 1.1µm solar reflectance 1.1 – 1.35µm Short Wave InfraRed 1.4 – 1.8µm solar reflectance (SWIR) 2 – 2.5µm Mid Wave 3 – 4µm reflectance, solar, thermal InfraRed (MWIR) 4.5 – 5µm temperature Thermal 8 – 9.5µm thermal temperature InfraRed (TIR) 10 – 14µm temperature (pas- thermal (passive) microwave, radar 1mm – 1m sive) artificial (active) roughness (active)
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Wien’s Law
• Specifies wavelength at which maximum BB radiation occurs
• Differentiate Planck’s equation, set to zero and solve for λ
λ = 2898 ⁄ T max
where λ is in µm and T is in K
λ • As T increases, |max decreases Stefan-Boltzmann’s Law
• Specifies total energy radiated by BB over all wavelengths
• Integrate Planck’s equation over all wavelengths
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5 4 2π k 4 M = ------T tot 2 3 -2 15c h (W-m ) 4 = σT
where σ = Stefan-Boltzmann constant =
5.67 x 10-8 (W-m-2-K-4) Examples for Wein’s and Stefan-Boltzmann’s Laws:
λ µ -2 source T (K) |max ( m) Mtot (W-m )
earth 300 9.66 (TIR) 4.6 x 102 incandescent 2800 1.04 (NIR) 3.5 x 105 lamp 0.483 (blue- sun 6000 7 green) 7.3 x 10
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Photometry
Radiometry in the context of Human Vision System (HVS) Luminous flux Φ in lumens (lm)
• Corresponds to radiometric flux in Watts (W)
• Incorporates the HVS sensitivity to radiation Conversion of radiometric units to photometric units
• Multiply spectral quantity of interest by photopic visual sensitivity curve
• Integrate over λ
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Ex: Find the effective visual lm-m-2 emitted by the sun
• Photopic visual sensitivity: V(λ), 0 ≤ V(λ) ≤ 1, unitless
• Scaling factor is 683 lm-W-1 at λ = 555nm (peak of V(λ))
solar M relative visual M photopic visual sensitivity 1.2 108 1
1 108 0.8
8 107 photopic visualsensitivity ) -1
m 0.6 µ - -2 6 107
0.4 solar M (w-m 4 107
0.2 2 107
0 0 400 450 500 550 600 650 700 wavelength (nm)
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• Spectral radiant exitance: MλΒΒ (T = 6000K, radiometric quantity)
• Total effective luminous exitance (photometric quantity)
V ()λ λ -2 Meff = ∫683V MλBBd (W-m )
–2 V × 9 • Using data in following table, Meff = 6.88 10 lm– m
Why isn’t this 1011?
Table for radiometric —> photometric conversion
-2 -1 -2 -1 λ (nm) Mλ (W•m •µm )V(λ)V(λ)•Mλ(W•m •µm ) 400 9.13e+07 4.00e-04 3.65e+04 410 9.34e+07 1.20e-03 1.12e+05 420 9.52e+07 4.00e-03 3.81e+05 430 9.67e+07 1.16e-02 1.12e+06 440 9.79e+07 2.30e-02 2.25e+06 450 9.88e+07 3.80e-02 3.75e+06 460 9.95e+07 6.00e-02 5.97e+06 470 9.99e+07 9.10e-02 9.09e+06 480 1.00e+08 1.39e-01 1.39e+07
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490 1.00e+08 2.08e-01 2.08e+07 500 9.98e+07 3.23e-01 3.22e+07 510 9.93e+07 5.03e-01 4.99e+07 520 9.88e+07 7.10e-01 7.01e+07 530 9.80e+07 8.62e-01 8.45e+07 540 9.72e+07 9.54e-01 9.27e+07 550 9.62e+07 9.95e-01 9.57e+07 560 9.52e+07 9.95e-01 9.47e+07 570 9.40e+07 9.52e-01 8.95e+07 580 9.28e+07 8.70e-01 8.07e+07 590 9.14e+07 7.57e-01 6.92e+07 600 9.01e+07 6.31e-01 5.69e+07 610 8.87e+07 5.03e-01 4.46e+07 620 8.72e+07 3.81e-01 3.32e+07 630 8.57e+07 2.65e-01 2.27e+07 640 8.42e+07 1.75e-01 1.47e+07 650 8.27e+07 1.07e-01 8.85e+06 660 8.11e+07 6.10e-02 4.95e+06 670 7.95e+07 3.20e-02 2.54e+06 680 7.80e+07 1.70e-02 1.33e+06 690 7.64e+07 8.20e-03 6.26e+05 700 7.48e+07 4.10e-03 3.07e+05
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Radiometric Transfer
Radiance
• Fundamental quantity for extended sources
• Flux per source area per solid angle radiated
• Measured normal to the source dA dΩ 2 ∂φ L = ∂Ω∂A propagation
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• Measured at an angle to the source • equivalent to a tilted source • use projected area θ
dΩ 2 ∂φ L = ∂Ω∂ θ propagation Acos dA
• NOTE: dΩ must be calculated for receiver surface normal to propagation direction (sometimes called projected solid angle) Inverse Square Law
• Point source emits radiation in all directions
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• Radiant intensity I (W-sr–1)
• Receiver normal to propagation:
d I dA
• Solid angle subtended by receiver at source
dA dΩ = ------2 d
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dA • Flux at receiver dΦ ==IdΩ I ------2 d
• Irradiance at receiver (W-m-2)
dΦ I dE ==------2 dA d
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Cos Law
• Tilt detector:
θ
I d dA
• Solid angle subtended by receiver at source
dAcosθ dΩ = ------2 d
• Irradiance at receiver (W-m-2)
dΦ I dE ==------cosθ 2 dA d
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Cos2 Law
• Move tilted detector to off-axis angle θ along receiver plane:
dA d/cosθ
θ I d
• Solid angle subtended by receiver at source
2 dA dA()cosθ dΩ = ------= ------2 2 []d ⁄ ()cosθ d
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• Irradiance at receiver (W-m-2)
dΦ I 2 dE ==------()cosθ 2 dA d
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Cos3 Law
• “Un-tilt” detector so that it’s in the receiver plane:
dA d/cosθ
θ I d
• Solid angle subtended by receiver at source
3 dAcosθ dA()cosθ dΩ = ------= ------2 2 []d ⁄ ()cosθ d
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• Irradiance at receiver (W-m-2)
dΦ I 3 dE ==------()cosθ 2 dA d
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Example: optical throughput between two surfaces
• Total flux through system is invariant to direction of propagation
• Case I: surfaces parallel to each other, and perpendicular to normal direction
Ω 21 Ω 12
A1 d A2
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Φ L = ------AΩ LA A Φ ==LA Ω ------1 2 12 1 12 2 d LA A Φ ==LA Ω ------2 1 21 2 21 2 d
• Case II: source and receiver surfaces not parallel to each other, or perpendicular to normal direction
Ω 21 Ω 12
A1 θ θ 1 d 2 A2
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θ A2 cos 2 Φ ==LA cosθ Ω LA cosθ ------12 1 1 12 1 1 2 d θ A1 cos 1 Φ ==LA cosθ Ω LA cosθ ------21 2 2 21 2 2 2 d
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