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Optics for Engineers Chapter 12

Charles A. DiMarzio Northeastern University

Apr. 2014

• Power is Proportional to

– Area of Aperture Stop

– Area of Field Stop

– “” of the Source ()

Apr. 2014 c C. DiMarzio (Based on for Engineers, CRC Press) slides12r1–1 Radiometry and

φ Radiant Radiant , M Note: “Spectral X:” Xν, Exitance, = W units /Hz or Xλ, /µm. W/m2 Rad. to Phot: , Luminous 683 lm/W × y (λ). Exitance, = 2 y (555nm) ≈ 1. lm lm/m = ∂/∂A → = lx

↓ ∂/∂Ω ↓ ↓ ∂/∂Ω ↓

← ·/R2 ∂/∂A → E , W/m2 , lm/m2 = lux I Radiant L Radiance, , W m2 sr W sr / / / , Luminous nit = Intensity, 2 lm/sr = cd lm/m /sr

1Lambert = 1lm/cm2/sr/π 1Ft.Candle = 1lm/ft2 1mLambert = 1lm/m2/sr/π 1FtLambert = 1lm/ft2/sr/π

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–2 Radiometric Quantities

Quantity Symbol Equation SI Units Q 3 = d Q Joules m3 w w dV 3 / or Power or Φ Φ = dQ W P dt = dΦ W m2 M M dA / Irradiance = dΦ W m2 E E dA / = dΦ W sr I I dΩ / Radiance = d2Φ W m2 L lm dA cos θdΩ / Fluence Ψ dQ J m2 dA / Fluence Rate dΨ J m2 F dt /   = M Dimensionless Mbb Spectral () () d() () Hz ν dν / or () d() () m λ dλ /µ Luminous Flux or Power P or Φ lm Luminous Exitance = dΦ lm m2 M M dA / Illuminance = dΦ lm m2 E E dA / = dΦ lm sr I I dΩ / Luminance = d2Φ lm m2 L L dA cos θdΩ / Spectral Luminous Eﬃciency V (λ) Dimensionless Color Matching Functions x¯ (λ) y¯(λ) z¯(λ) Dimensionless

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–3 Irradiance

• Poynting Vector for Coherent Wave dP dS ≈ (sin θ cos ζxˆ + sin θ sin ζyˆ + cos θzˆ) 4πr2 • Irradiance (Projected Area, Aproj = A cos θ) 2 d P ( dP ) dE = 0 = dA 4πr2

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–4 Irradiance and Radiant Intensity

A0 ·· Ω = r2 EI ·

• Intensity from Irradiance (Unresolved Source, A)

I I = Er2 E = r2

2 2 d P d P ( dP ) dI = = 0 = dΩ dA 4πr2 r2

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–5 Intensity and Radiance

• Resolved Source: Many Unresolved Sources ·· Combined ∫ ∫ · IL ∂I I = dI = dA ∂A • Radiance • On Axis (x0 = y0 = 0)

∂I (θ, ζ) ∂I (θ, ζ) ∂2P L (x, y, θ, ζ) = L (x, y, θ, ζ) = = ∂A ∂A ∂A∂Ω • Oﬀ Axis (Projected Area) 2 ∫ ∂I ∂ P L = = I (θ, ζ) = L (x, y, θ, ζ) dxdy ∂A cos θ ∂A∂Ω cos θ A

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–6 Radiant Intensity and Radiant Flux

• Solid Angle ·  √  Φ ( ) · I · NA 2 Ω = 2π 1 − 1 −  • Flux or Power, P or Φ n • Integrate Intensity

∫ ∫ P = Φ = I (θ, φ) sin θdθdζ

• Constant Intensity

P = Φ = IΩ ∫ ∫ 0 0 Ω = sin θ dθ dζ

∫ ∫ ( √ ) 2π Θ 0 0 Ω = sin θ dθ dζ = 2π (1 − cos Θ) = 2π 1 − 1 − sin2 Θ 0 0

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–7 Radiance and Radiant Exitance

· M ·· L • Radiant Exitance from a Source (Same Units as Irradiance) ∫ ∫ [ ] ∂2P M (x, y) = sin θdθdζ ∂A∂dΩ • Radiant Exitance from Radiance

∫ ∫ M (x, y) = L (x, y, θ, ζ) cos θ sin θdθdζ

∂M (x, y) 1 L (x, y, θ, ζ) = ∂Ω cos θ

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–8 Radiance and Radiant Exitance: Special Cases

• Constant L and Small Solid Angle

M (x, y) = LΩ cos θ  √  ( ) ( ) NA 2 NA 2 Ω = 2π 1 − 1 −  or Ω ≈ π n n

• Constant L, over Hemisphere ∫ ∫ 2π π/2 sin2 π M (x, y) = L cos θ sin θdθdζ = 2πL 2 . 0 0 2

M (x, y) = πL (Lambertian Source)

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–9 Radiant Exitance and Flux

Φ M ···

• Power or Flux from Radiant Exitance

∫ ∫ P = Φ = M (x, y) dxdy,

∂P M (x, y) = ∂A

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–10 The Radiance Theorem in Air

• Solid Angle two ways A0 A Ω = Ω0 = r2 r2

• Power Increment 0 dA 0 0 0dA dP = LdAdΩ = LdA cos θ dP = LdA dΩ = LdA cos θ r2 r2

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–11 Using the Radiance Theorem: Examples Later

• Radiance is Conserved in a Lossless System (in Air)

• Losses Are Multiplicative – Fresnel Reﬂections and Absorption

• Radiance Theorem Simpliﬁes Calculation of Detected Power – Determine Object Radiance

– Multiply by Scalar, Ttotal, for Loss – Find Exit Window (Of a Scene or a Pixel) – Find Exit Pupil – Compute Power

P = LobjectTtotalAexit windowΩexit pupil

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–12 and the Radiance Theorem

• Abbe Invariant: 0 0 0 n x dα = nxdα • Etendue ( ) 0 2 0 0 n2AΩ = n A Ω • Power Conservation ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 0 0 0 Ld2Ad2Ω = L d2A d2Ω

∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 0 ( ) L L 0 2 0 0 n2d2Ad2Ω = n d2A d2Ω n2 n2

L L0 Radiance Theorem: = n2 (n0)2

• Basic Radiance, L/n2, Conserved (Thermodynamics Later)

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–13 Radiance Theorem Example: Translation

• Translation Matrix Equation ( ) ( ) [( ) ( )] ( ) dx 1 z x + dx x dx 2 = 12 1 1 − 1 = 1 dα2 0 1 α1 + dα1 α1 dα1

dx2dα2 = dx1dα1

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–14 Radiance Theorem Example: Imaging

• Imaging Matrix Equation ( ) m 0 M 0 = SS ? n n0m

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–15 Radiance Theorem Example: Dielectric Interface

• Dielectric Interface Matrix Equation ( ) 1 0 0 n n0 • Power Increment 2 d Φ = L1dA cos θ1dΩ1 = L2dA cos θ2dΩ2

L1dA cos θ1 sin θ1dθ1dζ1 = L2dA cos θ2 sin θ2dθ2dζ2 • Snell’s Law and Derivative

n1 sin θ1 = n2 sin θ2 n1 cos θ1dθ1 = n2 cos θ2dθ2 sin θ n dθ n cos θ 1 = 2 1 = 2 2 sin θ2 n1 dθ2 n1 cos θ1 • Radiance Theorem L L0 = n2 (n0)2

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–16 Radiance Theorem Example: 1/2 Telecentric Relay

• Fourier–Optics Matrix Equation ( ) 0 M f FF 0 = 1 f 0

( )2 M 0 2 0 0 0 det FF 0 = 1 = n /n n AΩ = n A Ω • Radiance Theorem L L0 = n2 (n0)2

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–17 Idealized Example: 100W Lamp

• P = 100W, Uniform in Angle • Tungsten Filament: Area A = (5mm)2, Distance, R • Radiant Exitance

M = 100W/ (5mm)2 = 4W/mm2 • Radiance (Uniform in Angle)

L = M/ (4π) ≈ 0.32W/mm2/sr • Intensity I = 100W/ (4π) ≈ 7.9W/sr • Reciever: Area A0: Received Power and Irradiance ( ) 0 0 I PA / 4πr2 = IA /r2 E = ≈ 8W/sr/ (10m)2 = 0.08W/m2 r2

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–18 100W Lamp at the Receiver

A 0 E = L = LΩ r2

• Irradiance Same as Previous Page (5mm)2 E ≈ 0.32W/mm2/sr × ≈ 0.08W/m2 (10m)2

• Useful if r and A are Not Known

– Only Depends on Ω0 (Measured at Reciever)

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–19 Practical Example: Imaging the Moon

2 • Known Moon Radiance Lmoon = 13W/m /sr • Calculation – Jones matrices for Transmission (0.25), other multiplica- ( )12 tive losses (12 Lenses: 0.962 ).

L = 13W/m2/sr×0.25×0.9624 = 13W/m2/sr×0.25×0.367 = 1.2W/m2/sr – Exit Pupil (NA = 0.1) and exit window (Pixel: d = 10µm). – Irradiance, E = LΩ and Power on a Pixel, P = EA = LAΩ ( √ ) ( ) − 2 P = 1.2W/m2/sr × 2π 1 − 1 − NA2 × 10 × 10 6m

− − P = 1.2W/m2/sr × 0.315sr × 10 10m2 = 3.8 × 10 12W – If desired, multiply by time (1/30sec) to obtain energy. ∗ About One Million (and ) – Alternative to Solve in Object Space (Need Pixel Size on Moon)

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–20 Radiometry Summary

• Five Radiometric Quantities: Radiant Flux Φ or Power P , Radiant Exitance, M, Radiant Intensity, I, Radiance, L, and Irradiance, E, Related by Derivatives with Respect to Pro- jected Area, A cos θ and Solid Angle, Ω.

• Basic Radiance, L/n2, Conserved, with the Exception of Mul- tiplicative Factors.

• Power Calculated from and Field Of View in Image (or Object) Space, and the Radiance.

• Losses Are Multiplicative.

• Finally “Intensity” is Not “Irradiance.”

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–21 Spectral Radiometry Deﬁnitions

• Any Radiometric Quantity Resolved Spectrally

– Put the Word Spectral in Front

– Use a Subscript for or

– Modify Units

dL 2 dL 2 Lν = W/m /sr/THz or L = W/m /sr/µm dν λ dλ

• Spectral Fraction X (λ) f (λ) = λ for X = Φ,M,I, E, or L λ X

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–22 Spectral Radiometric Quantities

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–23 Spectral Fraction: on Earth

• 5000K (Discussed Later) Deﬁnes Mλ ∫ ∞ Mλ (λ) fλ (λ) = M = Mλdλ M 0

∗ • Compute Spectral Irradiance with Known E = 1000W/m2

Mλ (λ) Mλ (λ) Eλ (λ) = Efλ (λ) , with fλ (λ) = = ∫ ∞ M 0 Mλ (λ) dλ

In Practice: Use the spectral fraction with any radiometric quantity

* Solar Spectral Irradiance is 0 L Ω − E = λ e α dr λ r2 but we need to know a lot to compute it.

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–24 Spectral Fraction: Molecular Tag

400 0.01

• Excitation: Argon 488nm • Green Emission Power, Φ 200 0.005

Φλ (λ) = Φfλ (λ) , S (λ) f (λ) = ∫ λ λ ∞ 0 0 S (λ) dλ Absorption, Emission, \% 0 λ 300 400 500 600 700 – Dash–Dot on Plot λ, Wavelength, nm • Fluorescein Spectra – Obtained from – Dash: Absorption Experiment (Radiometric Calibration – Solid: Emission not Needed) Spectrum – Available from Vendor – Dash–Dot: Spectral Fraction for Emission Apr. 2014 c C. DiMarzio (Based on Optics for Engineers(nm−, CRC1) Press) slides12r1–25 Molecular Tag in Epi–Fluorescence

Narrow Filter Lost Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–26 Spectral Matching Factor (1)

400 0.01 18 Layers SMF = 0.67198 1 100

200 0.005 0.5 50 R, Reflectivity

0 Input and Output 0 0 0 400 600 800 400 600 800 Absorption, Emission, \% 300 400 500 600 700 λ, Wavelength, nm λ, Wavelength, nm λ, Wavelength, nm Fluorescein Emission Filter Spectral Matching Lλ (λ) R (λ) Lλ (λ) R (λ)

• Reﬂective Filter (Ch. 8) • Spectral Radiance Output of Filter 0 Lλ (λ) = Lλ (λ) R (λ) • Total Radiance Output ∫ ∞ ∫ ∞ 0 0 L = Lλ (λ) dλ = Lλ (λ) R (λ) dλ 0 0

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–27 Spectral Matching Factor (2)

• Total Radiance (Previous Page) ∫ ∞ ∫ ∞ 0 0 L = Lλ (λ) dλ = Lλ (λ) R (λ) dλ 0 0 • Deﬁne Reﬂectance for This Filter; R = L0/L – One Number as Opposed to R (λ) – Relate to Transmission at Spectral Maximum – Characterize SMF for Given Input Spectrum

∫ ∞ R (λ) R = Rmax fλ (λ) dλ = RmaxSMF 0 Rmax

R SMF = Rmax

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–28 Summary of Spectral Radiometry

• For every radiometric quantity, there exists a spectral radio- metric quantity. In the name, no distinction is made between frequency and wavelength derivatives.

• The notation: subscript ν for frequency or λ for wavelength.

• The units are the original units divided by frequency or wave- length units.

• Spectral Fraction, fλ can be applied to any of the radiometric quantities.

• The spatial derivatives (area and angle) are valid for the spectral quantities, wavelength–by–wavelength.

• The behavior of ﬁlters is more complicated, and is usually treated with the spectral matching factor.

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–29 Photometry and

• Spectral Luminous Eﬃciency, y¯(λ) • Source Spectral Radiance, 1.4

Lλ (λ, x, y) 1.2 • Eye Response 1 ∫ ∞ 0.8

= ¯ ybar Y (x, y) y (λ) Lλ (λ, x, y) dλ 0.6 0 • Four LEDs: Equal Radiance 0.4 – Blue, 400 Appears Weak 0.2 0 300 400 500 600 700 800 – Green, 550 Appears Strong Wavelength, nm. – Red, 630 Moderately Weak – IR, 980 Invisible

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–30 Lumens

• Power or Radiant Flux () ∫ ∞ P = Pλ (λ) 0

• Eye Response ∫ ∞ Y = y¯(λ) Pλ (λ) dλ 0

• Luminous Flux (Lumens, Subscript V for Clarity) ∫ 683 lumens/Watt ∞ P(V ) = y¯(λ) Pλ (λ) dλ max (y¯) 0

• Luminous Eﬃciency ∫ ∞ P( ) y¯(λ) P (λ) V = 683 lumens/Watt λ dλ P 0 max (y¯) P

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–31 Some Typical Radiance and Luminance Values

Object W/m2/sr nits = lm/m2/sr Footlamberts lm/W Minimum Visible 7 ×10−10 Green 5 ×10−7 1.5×10−7 683 Dark Clouds 0.2 Vis 40 12 190 Lunar disk 13 Vis 2500 730 190 Clear Sky 27 Vis 8000 2300 300 Bright Clouds 130 Vis 2.4×104 7 ×103 190 300 All 82 Solar disk 4.8×106 Vis 7 ×108 2.6×107 190 1.1×107 All ×107 82

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–32 Tristimulus Values: Three is Enough

• X, Y , Z 2 ∫ ∞

X = x¯ (λ) Lλ (λ) dλ 1.5 z 0 ∫ y x ∞ 1 Y = y¯(λ) Lλ (λ) dλ xbar,ybar,zbar ∫0 ∞ 0.5 x

Z = z¯(λ) Lλ (λ) dλ 0 0 • 300 400 500 600 700 800 Example: 3 Wavelength, nm. Krypton λred = 647.1nm 2 xbar ∫ ∞ ybar XR = x¯ (λ) δ (λ − 647.1nm) dλ =x ¯ (647.1nm) zbar 0 1.5

XR = 0.337 YR = 0.134 ZR = 0.000 Krypton λgreen = 530.9nm 1

XG = 0.171 YG = 0.831 ZG = 0.035 xbar,ybar,zbar

Argon λblue = 457.9nm 0.5

XB = 0.289 YB = 0.031 ZB = 1.696

0 300 400 500 600 700 800 Wavelength, nm. Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–33 Chromaticity Coordinates (1)

• Three Powers, R, G, and B, Watts • Tristimulus Values

      X XR XG XB R        Y  =  YR YG YB   G  Z ZR ZG ZB B

• Chromaticity Coordinates (Normalized X, Y

X Y x = y = X + Y + Z X + Y + Z

• Monochromatic Light x¯ (λ) y¯(λ) x = y = x¯ (λ) +y ¯(λ) +z ¯(λ) x¯ (λ) +y ¯(λ) +z ¯(λ)

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–34 Chromaticity Coordinates (2)

Monochromatic Light (Boundary) 1 2000K 3000K 0.8 3500K x¯ (λ) λ x = =457.9 nm x¯ (λ) +y ¯(λ) +z ¯(λ) λ=530.9 nm λ=647.1 nm 0.6 y¯(λ) RGB y = x¯ (λ) +y ¯(λ) +z ¯(λ) y 0.4 3 Lasers (Triangles) and Their Gamut (Black - -) 0.2 Phosphors (White) 0 Thermal Sources 0 0.2 0.4 0.6 0.8 1 (Later) x

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–35 Generating Colored Light

• • Powers of Three Lasers Given P(V ), x, and y • Required Tristimulus Values   R    G  = P( ) Y = V × B y 683lm/W  −1   XR XG XB X      YR YG YB   Y  ZR ZG ZB Z xP( ) X = V y × 683lm/W • Projector: 100lux “White” – x = y = 1/3 – 400lm on (2m)2 = 4m2 – X = Y = Z = (1 − x − y) P( ) Z = V 400lm/(1/3)/683lm/W y × 683lm/W – R = 1.2W, G = 0.50W, B = 0.33W Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–36 Another Example: Simulating Blue “Laser” Light

• Monochromatic Light from Blue Argon Laser Line λ = 488nm X = 0.0605 Y = 0.2112 Z = 0.5630

• Solution with Inverse Matrix R = −0.2429 G = 0.2811 B = 0.3261

• Old Bear–Hunters’ Saying:

– Sometimes You Eat the Bear, . . .

– and Sometimes the Bear Eats You.

• Best Compromise (Right Hue, Insuﬃcient Saturation) R = 0 G = 0.2811 B = 0.3261

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–37 Summary of Color

• Three color–matching functionsx ¯,y ¯, andz ¯ • Tristimulus Values: 3 Integrals X, Y , and Z Describe Visual Response • Y Related to Brightness. 683lm/W, Links Radiometric to Photometric Units • Chromaticity Coordinates, x and y, Describe Color. Gamut of Human Vision inside the Horseshoe Curve. • Any 3 sources R, G, B, Determine Tristimulus Values. • Sources with Diﬀerent Spectra May Appear Identical. • Light Reﬂected or Scattered May Appear Diﬀerent Under Diﬀerent Sources Of Illumination. • R, G, and B, to X, Y , Z Conversion Can Be Inverted over Limited Gamut.

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–38 The Radiometer or

• Aperture Stop • Calibration Required • Field Stop • Spectrometer on Output? • Measured Power • Spectral Filter? 0 0 • Photometric Filter? P = LAΩ = LA Ω • Computer • Adjustable Stops – Radiance,P/(ΩA) (Match FOV) – Luminance (All Units) • Sighting Scope? – Spectral Quantities

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–39 Integrating Sphere

• Power from Intensity • Integrate over Solid Angle – Goniometry ∗ Information–Rich ∗ Time–Intensive ∗ Integrating Sphere · Easy · Single Measurement • Applications – Wide–Angle Sources – Diﬀuse Materials • Variations – Two Spheres – Spectroscopic Detector – More

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–40 Blackbody Outline

• Background • Equations, Approximations • Examples • Illumination • Thermal Imaging • Polar Bears, Greenhouses

10 10 10000

m 5000 µ 8 / 2 10 3000 1000 6 500 10 300 100 4 10

2 10

0

, Spect. Radiant Exitance, W/m 10 λ m

−2

10 −2 −1 0 1 2 10 10 10 10 10 λ, Wavelength, µ m Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–41 Blackbody Background

• Cavity Modes

2πνx` = Nxπ c

2πνx` = Nyπ c

2πνx` = Nzπ c • Mode √ 2 2 2 c ν = ν + ν + ν ν0 = • Waves in a Cavity x y z 2` ∂2E ∂2E ∂2E 1 ∂2E + + = √ 2 2 2 2 2 2 2 2 ∂x ∂y ∂z c ∂t ν = ν0 Nx + Ny + Nz

• Boundary Conditions 2πνxx 2πνyy 2πνzz E (x, y, z, t) = sin sin sin sin 2πνt c c c Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–42 Counting Modes

• × 2 for Polarization Temperature = 5000K 8πν2`3 Nν = c3 • Energy Distribution – Rayleigh–Jeans: Equal Energy, kBT (Dash–Dot) −20 8πν2`3 10 = 0 Jν kBT 3 Q, Energy Density, J/Hz 10 c λ, Wavelength, µ m ∗ ”Ultraviolet • Mode Density (Sphere) Catastrophe” ∗ Wein exp ( ) 4πν2dν/8 hν hν/kT dN = per Mode c3/(8`3) 8πν2`3 dN 4πν2`3 Jν = hν exp (hν/kT ) = = c3 Nν 3 dν c · Dashed

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–43 Planck Got it Right

Temperature = 5000K • Boltzmann Distribution − JN k T P (JN ) = e B • Average Energy

∑∞ −Nhν Nhνe kBT ¯ = N=0 = J ∑∞ −Nhν −20 e kBT 10 N=0 0 ∑ − Q, Energy Density, J/Hz 10 ∞ Nhν Ne kBT λ, Wavelength, µ m N=0 hν ∑∞ −Nhν e kBT • Planck N=0 • With Some Work. . . – Mode Energy Quantized ehν/kBT 2 J = Nhν ¯ (1−ehν/kBT ) N J = hν 1 = – N Random (1−ehν/kBT ) – Average Energy Density hν ∑ ∞ J P (J ) ehν/kBT − 1 ¯ N=0 N N J = ∑∞ N=0 P (JN ) Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–44 Planck’s Result (1)

• • Previous Page 2. Emssion Balanced by Ab- sorption hν hν/k T ¯ ¯ e B − 1 M(T )emitted = aE(T )incident • × Mode Density • 3. Balance at Every λ 8πν2`3 hν • 4. Light Moves at c N J¯ = ν 3 hν/k T ∫ c e B − 1 1 4πl • wλ = LdΩ = Solid Curve (Previous Page) c sphere c • Energy Per Volume • E from L 2 ∫ ∫ dw¯ NνJ¯ 8πν hν = = Eλ = Lλ cos θ sin θdθdζ = dν `3 c3 ehν/kBT − 1 • 1. Radiance Independent of Direction πLλ

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–45 Planck’s Result (2)

• E from w: Eλ = 4cwλ • ¯ ¯ Wall Balance & a = 1: M(T )bb = E(T )incident

• Planck’s Law 2πν2 1 2πhν3 1 Mν(bb) (λ, T ) = hνc = c3 ehν/kBT − 1 c2 ehν/kBT − 1 • Emissivity

Mν (λ, T ) = Mν(bb) (λ, T ) • Detailed Balance

 (λ) = a (λ)

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–46 The Planck Equation

• Fractional Linewidth

|dν/ν| = |dλ/λ| dM dν dM ν dM c dM M = = = = λ dλ dλ dν λ dν λ2 dν

• Planck Law vs. Wavelength

2πhc2 1 Mλ (λ, T ) = λ5 ehν/kBT − 1

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–47 Useful Blackbody Equations

• Wien Displacement 10 Law 10 10000

m 5000 µ 8 /

2 10 · 3000 λpeakT = 2898µm K 1000 6 500 10 300 • Stefan–Boltzmann 100 4 10 Law

2 10

5 4 4 0

2π k T , Spect. Radiant Exitance, W/m 10 M (T ) = λ 3 2 m 15h c −2

4 10 −2 −1 0 1 2 = σeT 10 10 10 10 10 λ, Wavelength, µ m

• Stefan– −12 2 4 −8 2 4 σe = 5.67032 × 10 W/cm /K = 5.67032 × 10 W/m /K

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–48 Some Examples: Thermal Equilibrium

• Body Temperature • Earth Temperature – Cooling (310K) – Heating 4 2 M (T ) = σeT = 524W/m ∗ 1kw/m2 ∗ Half of Surface – Heating (Room=295K) – Cooling 4 2 M (T ) = σeT = 429W/m ∗ Radiation – Excess Cooling 4 M (T ) = σeT ∆M × A ≈ 100W – Result ≈ 2000kCal/day T ≈ 306K (Many Approximations)

( ) Because d T 4 = 4T 3dT , small changes in temperature have big eﬀects on heating or cooling.

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–49 Temperature and Color

• Radiant Exitance ∫ ∞ • Tristimulus Values M(R) = M(R)λdλ     0 X ∫ ∞ x¯ • Luminous Exitance      Y  =  y¯ M(R)λdλ ∫ 0 ∞ Z z¯ M(V ) = M(V )λdλ 0 • Integrands for × 2000 3000, 3500K M(V ) = 683lm/W 5 ∫ ∞ 10 4 yM¯ dλ 10 (R)λ xs,ys,zs 3 0 10 300 400 500 600 700 800 λ, Wavelength, nm 7 10 6 10 10 10 10 10 Solar Disk xs,ys,zs 5

Solar Disk /sr 10 /sr 2

2 300 400 500 600 700 800 Lunar Disk λ, Wavelength, nm 7 Lunar Disk 10 0 0 10 10 6 10

Min. Vis. xs,ys,zs 5 M, Radiance, W/m

M, Luminance, L/m 10 300 400 500 600 700 800 Min. Vis. −10 −10 λ, Wavelength, nm 10 10 0 2000 4000 6000 8000 10000 Apr. 2014 T, Temperature, c C. DiMarzio K (Based on Optics for Engineers, CRC Press) slides12r1–50 Chromaticity Coordinates of Blackbody

• Chromaticity

X 1 x = 2000K X + Y + Z 3000K 0.8 3500K Y λ=457.9 nm y = λ=530.9 nm λ=647.1 nm X + Y + Z 0.6 RGB

0.9 y x y 0.8 0.4

0.7

0.6 0.2 0.5

0.4 x,y, Color Coordinates 0 0.3 0 0.2 0.4 0.6 0.8 1 x 0.2 0 2000 4000 6000 8000 10000 T, Temperature, K

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–51 Solar Spectrum

• Exo–Atmospheric • Sea Level – 6000K, 1480W/m2 – 5000K, 1000W/m2 2 2 Eλ (λ) = 1560W/m × Eλ (λ) = 1250W/m ×

fλ (λ,6000K) fλ (λ,5000K)

2500 2000

2000 1500 m m µ µ / 1500 / 2 2 1000 1000 , W/m , W/m λ λ e e 500 500

0 0 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 2.5 3 3.5 4 λ, Wavelength, µ m λ, Wavelength, µ m

Constants are higher than total irradiance to account for absorption in certain regions of the spectrum.

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–52 Outdoor Radiance

• × −4 • Sunlit Cloud Blue Sky: Mλ = Mskyfλ (5000K) λ

10 10 Eincident = m Cloud µ 8 / 2 10 Blue Sky 2 1000W/m Night 6 10 5000K 4 10 – White Cloud 2 10 Mcloud = Eincident 0

, Spect. Radiant Exitance, W/m 10 λ

– Lambertian m −2

10 −2 −1 0 1 2 = 10 10 10 10 10 Lcloud Mcloud/π λ, Wavelength, µ m

• Radiant Exitance, M of Object Surface Illuminated with E

Mλ (λ) = R (λ) Eλ (λ) Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–53 Thermal Illumination

• Thermal Sources (e.g. Tungsten) ∫ ∞ ∫ ∞

M(R) = M(R)λdλ M(V ) = 683lm/W yM¯ (R)λdλ 0 0

• Luminous Eﬃciency ∫ ∞ yM¯ dλ ∫0 (R)λ M(V )/M(R) = 683lm/W ∞ 0 M(R)λdλ

100 • Highest Eﬃciency 50 λ = 555nm

• Efficiency, lm/W Bad Color 0 0 2000 4000 6000 8000 10000 T, Temperature, K

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–54 Illumination

Thanks to Dr. Joseph F. Hetherington

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–55 IR Thermal Imaging

T = 300 K • Infrared Imaging 1 ∫

/Delta T 0.5 λ2 λ M ∆ ρ (λ)  (λ) Mλ (λ, T ) dλ 0 −1 0 1 2 λ1 10 10 10 10 – is Responsivity T = 500 K ρ 6

–  is Emissivity 4 /Delta T λ 2 • Maximize Sensitivity M ∆

0 −1 0 1 2 10 10 10 10 ∂Mλ (λ, T ) λ, Wavelength, µ m

10 ∂T 10 Sun

• m Cloud Work in Atmospheric Pass µ 8 / 2 10 Blue Sky Bands Night 6 10 • Shorter Wavelength for Higher 4 10 Temperatures 2 10 • Hard to Calibrate (Emissivity, 0

, Spect. Radiant Exitance, W/m 10 λ

etc.) m −2

10 −2 −1 0 1 2 • Reﬂectance and Emission 10 10 10 10 10 λ, Wavelength, µ m Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–56 Polar Bears

• Thermal Equilibrium E = 50W/m2, 200W/m2 – Heating by Sun incident 600W/m2, top to bottom ∗ High Temperature 80

∗ m 60

/ 40 2 20 0 ∗ Small Solid Angle −20 W/m −40 0 5 10 15 20 – Cooling to λ, Wavelength, µ m Surroundings 200 λ 100 M ∗ Body Temperature 0 0 5 10 15 20 ∗ Low Radiance λ, Wavelength, µ m ∗ 1000 Ω = 2π λ 500 M – Extra Heat from 0 0 5 10 15 20 Metabolism λ, Wavelength, µ m • Short–Pass Filter Heating (—), Cooling (- -), Net (··) – Pass Visible

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–57 Wavelength Filtering

1500 • Perfect Bear Net Cooling for Two Best Bare Bear 60 800nm LongPass 2.5µm ShortPass µ 1000 2.5 m LongPass Dynamic Filter Best Bear 50 2

40 2

, W/m 500 net M

, W/m 30 lost

0 M 20

10 −500 0 500 2 1000 1500 E , W/m 0 sun 0 500 1000 1500 E , W/m2 • Bare Bear: No Filter sun • • 800nm Short Pass Dynamic Filter Follows • 2.5nm Short Pass Like Zero Crossing of Net Mλ • Glass Perfect: No Cooling • Best Bear is a Dynamic (Impossible: Nothing Out Filter Means Nothing In)

Apr. 2014 c C. DiMarzio (Based on Optics for Engineers, CRC Press) slides12r1–58