Black Body Radiation and Radiometric Parameters:

 All materials absorb and emit radiation to some extent. A blackbody is an idealization of how materials emit and absorb radiation. It can be used as a reference for real source properties.  An ideal blackbody absorbs all incident radiation and does not reflect. This is true at all wavelengths and angles of incidence.  Thermodynamic principals dictates that the BB must also radiate at all ’s and angles.  The basic properties of a BB can be summarized as: 1. Perfect absorber/emitter at all ’s and angles of emission/incidence. Cavity

BB

2. The total radiant energy emitted is only a function of the BB temperature. 3. Emits the maximum possible radiant energy from a body at a given temperature. 4. The BB radiation field does not depend on the shape of the cavity. The radiation field must be homogeneous and isotropic.

T

If the radiation going from a BB of one shape to another (both at the same T) were different it would cause a cooling or heating of one or the other cavity. This would violate the 1st Law of Thermodynamics.

T T

A B Radiometric Parameters: 1. Solid Angle dA d  r 2 where dA is the surface area of a segment of a sphere surrounding a point.

r

d A r is the distance from the point on the source to the sphere. The solid angle looks like a cone with a spherical cap.

z

 r d

r r sind y

 r sin

x

An element of area of a sphere

2 dA rsin d d Therefore

dd  sin d The full solid angle surrounding a point source is:

2 dd  sind  00  2cos 0  4 Or integrating to other angles  < :

 21cos     The unit of solid angle is steradian.

2. Radiant Flux  and Energy Density: Radiant energy Q (J); Energy flux is the rate of radiant energy transferred from one point or surface.

Energy flux (Φ) is measured in watts. Can be spectral  or a total over all wavelengths. (Note that Flux is the same as optical power.) The energy density (u) is the energy per unit volume.   dQ/ dt udQdV / Note that Φ is the total flux integrated over all wavelengths. It consists of the integrated spectral flux or power with:

   d 0    d 0

Note that   but  d dsince this represents an equal increment of power. The scaling factor between units is found by using the relation:   c / c dd   2 c    2 3. Irradiance: The power per unit area illuminating a collection or detection surface.

d E  dAdet

4. Spectral Intensity: The power emitted per unit solid angle from a source. (Wavelength dependent)

d() I   d where dω is an increment of solid angle. Note that in physics intensity refers to the magnitude of the Poynting vector of an EM field. This interpretation resembles irradiance as defined here. Note that if we consider a receiving surface element dA, the solid angle subtended by this surface relative to a point source is ddAr cos / 2 . In this case the irradiance is related to the intensity by: d E  dA d I  2 cosdA / r I cos E r 2

5. Spectral Emitance: The power per unit source area emitted from a source. (Wavelength dependent.)

d() M   dAsrc

6. Spectral Radiance: The power emitted per unit of projected source area per solid angle. (Wavelength dependent.)

dI    L  cosdAsrc dcos dAsrc

L

d



dA 

Source n Area

Spectral intensity, emittance, and radiance are terms referring to an optical source. They can also be integrated over all wavelengths to obtain the total intensity, emittance, and radiance from the source. Radiance is the most general of the source radiometric terms.

The radiant energy dQ(λ) emitted from an area dA over time dt, wavelength interval d, and solid angle d in the direction ,  is related to the monochromatic radiance by:

dQ  Lcos dAd d dt . Note that this represents the energy emitted from the source element dA at an angle  relative to the surface normal nˆ . cos dA is the projected area of the source. The optical flux or power emitted from an element of source area dA can be estimated using the approximate relation:

 LA cos   . Q     where dt A Lambertian source emits light with the following characteristic: I()  Ld cos A  o src , II() o cos I  LdA with oo . src

Io

Iocos  

A BB acts as a Lambertian emitter. For a BB the emittance and radiance are related according to: MLcos d   /2  2sincos L d   0 1  2coscos Ld.    0

  L Note that the definition of the solid angle: dd  sin is usedd with

02  . Relations between Parameters:  From a point source: IW(/) sr  4 At a distance r from the point source the irradiance on a plane perpendicular to r is: I  E  . rr224

The irradiance was obtained by integrating over the area surrounding a point that emits power . It can be seen that the power decreases as 1/r2 which is a well known radiation property. Even if a source has a finite diameter it is still possible to use the inverse- square distance relation to approximate the power emitted from a source

Source

Observer

S D provided that the ratio of the source diameter-to distance is small enough.

Src distance/diameter Subtended Angle Inverse Sq. Law Error ratio 1.6 ~32o ~10% 5 ~11.3o ~1% 10 ~5.7o ~0.25% 16 ~3.6o ~0.1%

Note that the limit of angular resolution for the human eye is ~ 0.1o. Therefore for visual applications a ratio of distance/source diameter of 16 is adequate for approximating an extended source as a point source. The angle subtended by the sun is 0.52o therefore it is definitely a point source when viewed from the Earth. Conservation of Radiance: The radiance theorem is an important law of radiometry and states that radiance is conserved with propagation through a lossless optical system.

The radiance measured at the source and receiver is compared.

Source Receiver

dAo r dA 1 o 1

Consider a source area dAo and receiver area dA1 separated by a distance r.

The corresponding solid angles are:

do = the solid angle subtended by dA1 at dAo

dA cos d 11 o r 2

d1 = the solid angle subtended by dAo at dA1

dA cos d 00 1 r 2

If L0 is the radiance of the radiation field measured at dA0 in the direction of dA1, the flux transferred from dA0 to dA1 is:

2 dLdAd 00 cos 0  0

Similarly the flux transferred from dA1 to dA0 is:

2 dLdAd 11 cos 1  1 .

The radiance L1 measured at dA1 is therefore:

d 2 L1  dA11cos d1

And since the flux originates from dAo: d 2 L dAcos d L 00 0 0 1 dAcos d dA cos d 111111 

LL10

This implies that the radiance of the source is the same regardless of where it is measured and is conserved.

Another interesting result can be obtained by re-writing the expression: 2 dA10cos dLdAd00cos 0  0  LdA 00 cos 02 r

LdA01cos 1 d 1

The result shows that the transmitted flux can be obtained by taking the projected area and solid angle product at the source or receiver.

This allows using a perspective either from the source or receiver to perform an analysis.

Lambertian Disk Source: It is desired to compute the irradiance from a disk Lambertian source of radius R and uniform radiance L at an area element dA1 that is parallel to the surface of the disk and is located axially at a distance z from the center of the disk.

An annular area element on the surface on the source is given by:

sind dA 2 z2 0 cos3  The element of solid angle subtended by dA1 from any point on dA0 is given by:

dA1 cos d0 2 z /cos The flux transferred from dA0 to dA1 is given according to the definition of radiance as:

2 dLdA2sincos 1 d

The irradiance at dA1 becomes:

d 1/2 E  2sincos Ld dA1 0 2 2 R LLsin 1/2 22 R  z

with

R 1  1/2  tan  z Note that when z << R the irradiance approaches a value of L. This is the same value found previously for the radiant exitance of a Lambertian source.

When z >> R the irradiance approaches

22 ERL  / z in this case the irradiance observes an inverse square law.

At very large z the source looks like a point source with the intensity of

2 IR  L and the irradiance for a point source is: R22 RL EL 22 2 R  zz I  z2 which illustrates the inverse square law dependence.

Example: Spherical Lambertian Source: Consider the irradiance on an area dA located a distance r from the center of a spherical Lambertian source with uniform radiance L. To determine this we will use the symmetry of the situation rather than integrating over the surface of the source.

R dA r

The radiant exitance of a Lambertian source is given by:

M   L

The total flux emitted by the source is

22 4 RL

At a distance r from the center of the source it radiates uniformly over an area 4 r 2 .

Therefore the irradiance at a distance r is given by:

 R2 L E  r 2 . The irradiance is seen to follow an inverse square law at a distance r. The corresponding intensity becomes:

 IR 2 L 4 .

An observer at dA looking back at the source will see a uniform disk with half angle:

1 R 1/2  sin  r .

Planck’s Radiation Law: The spectral radiant exitance from a BB is given by:

21hW3  M   22 chkTmexp / 1 Hz

21 hc2  W M   52 exphc / kT 1  m m

Note that due to the relation between  and 

M  M  but MdMd   c   

c dd    2

Stefan-Boltzmann Law: The total emittance from a BB is only a function of the BB temperature. It can be found by integrating M  over all wavelengths.

4 2   mWTM )/( The Stefan-Boltzmann constant:

54 2 kW8   s 235.673 10  2 4 15ch m K Boltzmann Constant: kJ1.38 1023 / K

Emissivity:   A BB is a perfect emitter with unity emissivity (ε =1).

 All practical bodies have  < 1. The curve below shows the difference between a BB and a non-ideal source.  The emissivity for a non-ideal source can be obtained through the following experiment.

 Consider an object that is illuminated with incident radiation o  From Energy Conservation:

RT r  1 where  R  1 ; 0

3 Tr  ; 0    2 0

 (BB)

S  

 The above figure shows the spectrum of a BB and a non-ideal source. Object

Incident 3

0 Transmitted

1 Reflected Absorbed  2  The emissivity can be shown to be equal to the absorption through the use of the Kirchoff Radiation Law.  Consider: An object at temperature T surrounded by an enclosure also at temperature T.

Enclosure Temp (T) Object Temp (T)

 At thermal equilibrium:

Emitted Absorbed . by Object by Object

 If the total flux incident on the object is 0 then the

Absorbed Flux  0 = Emitted Flux  0 and    . Therefore the emissivity coefficient (ε) can be obtained from the absorption coefficient (α) and using

RT r  1 ,  1RTr and for an opaque object

 1. R The emissivity will in general be wavelength dependent which would require evaluating the reflectance at each wavelength with a spectrometer. However over a limited spectral range it can be considered a constant.  Once the emissivity is found the spectral emitance of a non-ideal (BB)

source can be found by multiplying M by  to account for non-ideal emitters.

Wien’s Displacement LAW:

The wavelength at which the Planck function is a maximum M times the temperature of the corresponding BB is a constant.

M Tm 2892 K

The figure above shows the relative power emitted from a thermal source at T = 2000oK, 2250oK, and 2500oK as a function of wavelength. The peak wavelength can be predicted using the Wein displacement law. Example 1: First order radiometric properties of the sun. Peak solar radiation appears at   0.48m . Therefore

2892mK o TK6025o . 0.48m The sun appears as a blackbody source with a temperature near 6000oK. The peak radiant exitance can be computed: 3.74 108 M  0.48 m 4 5  1.44 10  0.48 exp 1  0.48 6025  1082Wm /  m

The radiance of the sun at 0.48 m using M    L :

8 10 2 LWmsrpeak  / m (Lambertian Assumption)   The spectral flux from the sun collected on the Earth: LdcosdA 0.488 mp eako A  sun d is the angle subtended by the Earth relative to the sun, and dA is an increment of area on the sun’s surface. Since the sun is a sphere it appears as a full disk to an observer and the observer appears normal to it. Therefore

oo0cos 1 Note: Expanding the product of the differentials, dA dA dA d dAcol sun dA d dA sun col sun rr22col sun col

2 6 Using a collection area Acol 1m and r = 9310 miles, Dsun = 0.865 million miles, then the angle subtended by the sun is 0.865/ 2 2    6.8 105 sr . sun 932 The flux from the sun hitting 1 square meter on Earth:

0.48 m  L  A  2 sun sun col Am1 peak 108  6.5 1052sr 1 m    2160Wm /

The solar output over all wavelengths: W MT 4 7.35 107 at T = 6000oK m2 M L  for a Lambertian emitter. 

7 7.35 10 52 2  6.8 10srm 1 Am1  1590W for a 1 m2 collector. Example 2: Flux collected by a detector on a surface. oo2 CollectedEAEc icos(30 ) Collector i cos(30 )(1m ) I(30o ) IE,, E  44rcm22i 20

oo II30 o cos  30 1WW ILA 11 cm2 oosourcecm2  str str

1/Wstr cos(30)o Ei 20cm 2 1cos(30)22o  1cm Collected 20cm 2 0.00188WmW 2