Chapter 2 Flashcards

ω λ wavelength f/v frequency μm micrometer τ transmissivity ξ absorptivity θ theta

1
Q

EMR defined

A
  • Form of energy transport in free space

* wave travel through space at speed of light

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2
Q

Maxwell equations

A

Esubx = Eomega cos (wt-kz)

Eomega = max electrical energy
ω= angular freq (2 piv)
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3
Q

Electric and magnetic fields are ________ to each other

A

Orthagonal (perpendicular?)

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4
Q

Radiant energy (Q of a photon) is proportional to frequency

A

Q=hv

Q=radiant energy (J)
v=frequency
h=Plank’s constant (6.626 X 10^-34 Js

substituteQ=hc/wavelength
h is small number, so approx Q~ 1/wavelength

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5
Q

Radiation from the Sun is ________________ (re polarization)

A

un-polarized

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6
Q

Man-made sources (laser, radar) have ___________ radiation (re polarization)

A

polarized

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7
Q

Electromagnetic spectrum for remote sensing

A

UV, visible, near-ir, mid-ir, thermal, microwave bands (Ka, Ku, X, C, S, L, P)–LOOK AT GRAPH FROM LECTURE

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8
Q

shorthand ranges for RGB spectra

A

blue .4-.5 μm
green .5-.6 μm
red .6-.7 μm

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9
Q

Polarization

A

The orientation of the electric field

Vertical polarization: electric vector is PERPENDICULAR to the plane of incidence

Horizontal polarization: electric vector is PARALLEL to the plane of incidence

Sun=unpolarized
Man-made sources (laser, radar) have polarized radiation

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10
Q

Infrared

A

near and mid=reflective =short wave

far=emissive radiative thermal

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11
Q

Microwaves

A

letters come from military, don’t want enemy to know signal

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12
Q

The bulk of sun’s radient energy distribution is

A

visible (43.5%) and near infrared (36.8%). Also significant amts of near UV (5.32%) and Mid IR (12%)

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13
Q

Blackbody concept

A

Object that absorbs and emits 100% of radiation
Does not exist in nature
Emissivity would equal 1

Assumptions:

  • Isotropic
  • homegeneous
  • unpolarized

2 objects with same temp would emit same E

Blackbody would emit more E than a comparable gray body (which has Emissivity is < 1)

MOST IMPORT. CONCEPT

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14
Q

Graybody

A

Object that reflects part of the incident radiant
Emissivity is < 1

M=emissivitysigma constantTemp4 (double check this)

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15
Q

Emissivity

A

the relative ability of a surface to emit radiation

  • describes ACTUAL absorption and emission properties of real objects (gray bodies)
  • Is wavelength dependent (usually use avg)
  • Is equal to (graybody emittance)/(blackbody emittance)–of same temp
  • Use it to calculate an object’s radiant temp, or brightness temp

The temp at which a blackbody would have to be to emit the same energy as emitted a graybody at some physical temp

T(rad) = e^(1/4)T(kin) e= [(T(rad) / T(kin)] ^4

kin=kinematic physical

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16
Q

Selective radiator

A

emits certain types of EMR (better)

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17
Q

Two objects can have the same ___________ temp but different _________ temperatures…Why?

A

kinematic, radiant

Because they have different emissitivities

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18
Q

Understand why mirror has no emissivity

A

gah

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19
Q

Planck’s law (Spectral radiance)

A

‘All bodies who temp are above absolute zero K (-273.2), emit radiation”

Heat transformed into radiant energy

(need formula)

L = amount of E per unit serface per unit time, per solid angle emitted at the wavelength λ

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20
Q

Maximum radiation of sun at what wavelength?

A

6000K (Kelvin) at ~.5 micrometer

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21
Q

Stefan-Bolzmann law (TIR)

A

The total emitted energy over the whole spectrum is proportional to the physical temperature.

The amount of energy emitted by an object such as the Sun or Earth is a function of its temperature. ^temp=^emitted energy

22
Q

Wien’s law

A

That wavelength of peak emittance (max wavelength) is inversely proportional to an object’s kinematic temperature (derivative of Planck’s law)

λmax = a/T

a=2898 μm K
e.g. hotter the object, the shorter wavelength of maximum intensity

23
Q

False color

A

Different channels represented with color not true to visible light. May more clearly delineate areas

24
Q

Solid angle

A

imagine as angle with three dimensions (cone)

25
Q

Look over definition of radiation quantities

A

tba (know what is what)

26
Q

EMR interaction with matter

A

Radiative properties of a natural surface
• Radiation incident upon a surface must either be transmitted (t) through it, reflected (a) from the surface or be absorbed (ξ).

Transmissivity (τsubλ) + Reflectivity (αsubλ) + Absorptivity (ξsubλ) = 1

  • For solar radiation, a is referred to as the surface albedo
  • If we consider only part of the EM spectrum, a is referred to as the spectral albedo
27
Q

albedo

A
  • For solar radiation, Absorptivity (a) is referred to as the surface albedo
  • If we consider only part of the EM spectrum, a is referred to as the spectral albedo
28
Q

transmission

A

Transmission
– incident radiation passes through the material without attenuation
– change in the direction of radiation is given by the index of refraction of the material
– Index of refraction (n) is the ratio of the speed of light in a vacuum relative to the speed of light through the material

n = c/c subn

Snell’s Law describes relationship between incidence and refraction angles:
Therefore
n1 sinq1 = n2 sinq2 n1/n2 = sinθ2/sinθ1

29
Q

Snell’s Law

A

Snell’s Law describes relationship between incidence and refraction angles:

n1 sinθ1 = n2 sinθ2

Therefore

n1/n2 = sinθ2/sinθ1

30
Q

Reflection

A

Four kinds kinds (all can happen at same time):

• Reflection (Specular Reflection)
– Surface is smooth relative to wavelengths (smooth relative to the wavelength)
– Mirror-like surfaces are called specular reflectors

• Scattering (Diffuse Reflection)
– Surface is rough relative to wavelengths
– EMR velocity and wavelength are not affected

  • Absorption
  • Attenuation
31
Q

Absorption

A

– Substance is opaque to incident radiation

– Portion of EMR is converted to heat energy (re-radiated)

32
Q

Attenuation

A

Weakening of EMR as it passes through a medium. Sometimes called extinction.
– Combination of absorption and scattering

33
Q

Plants absorb/reflect what

A

absorb blue and red, reflect green and infrared

34
Q

Water absorbs/reflects what?

A

reflects blue, green, red, IR, in progressively less amounts, until absorption at approx 800 nm

35
Q

Atmospheric Effects

A

• EMR is attenuated by its passage through atmosphere
• Attenuation= scattering + absorption
- Scattering is the redirection of radiation by reflection and refraction
- Attenuation is wavelength dependent
- Decrease with increase in wavelength

36
Q

Three types of scattering in the atmosphere

A

Rayliegh, Mie, and non-sective scattering

The type of scattering is a function of

1) wavelength
2) size of the gas molecule, dust particle, and/or water vapor droplet encountered

37
Q

Rayleigh scattering

A

(molecular scattering)
• Scattering by molecules and particles whose diameters are &laquo_space;wavelength
• Primarily due to oxygen (O2) and nitrogen (N2) molecules
• Scattering intensity is proportional to λ-4
• Responsible for blue sky (blue is shorter than other)

38
Q

Mie scattering

A

• Particles that have a mean diameter of 0.1 to ten times the incident wavelength
– Examples: water vapor, smoke particles, fine dust
– Scattering intensity is proportional to anywhere from λ-4 to
λsup0 (depending on particle size)
• Clear (all colors) atmosphere has both Rayleigh and Mie scattering. Their combined influence is between λ-0.7 and λ-2

Why we see red if there’s a fire or red sky–Mie scattering is favoring the longer wavelengths.

Red sky = angle in atmosphere means Rayleigh scatters more, so blue reflected into space.

39
Q

Non-selective scattering

A
  • All visible spectra scattered (then will have white/gray)
    • Aerosol particles much larger than wavelength (>10x)
    • Examples: water droplets, ice crystals, volcanic ash, smog
    • Independent of wavelength λ0
40
Q

Why does Rayleigh scattering make the sky blue?

A

Blue wavelength is shorter, so the blue light is scattered 4x more than red. Thus, the intensity of the blue is larger than the red (divided by smaller number).

41
Q

Why does Mie scattering make the sky red?

A

Mie scattering is favoring the longer wavelengths.

42
Q

What is difference between white and gray

A

more or less reflection/absorption. White scatters all of visible light. Gray reflects some of all visible light.

43
Q

Atmospheric Absorption

A
• In the atmosphere, EM radiation is absorbed by:
– H2O Water vapor, Water droplets
– CO2 Carbon dioxide
– O2 Oxygen (not very effective)
– O3 Ozone

• This absorption is the transfer of electromagnetic energy to the molecules with which the EMR comes in contact with

44
Q

Vibrational Processes (how absorption occurs)

A
  • Small displacements of atoms from their equilibrium position, resulting from absorption of energy
  • N atoms –> 3N possible vibrational modes
  • Figures show three modes for the water molecule (symmetric OH stretch, assymetric OH stretch, HOH bend)
45
Q

Vibrational Processes of water (three classical frequencies)

A

The water molecule has three classical frequencies (v1, v2, v3) which correspond to three wavelengths:

  • 3.106 μm (symmetric OH stretch)
  • 6.08 μm (HOH bend)
  • 2.903 μm (asymmetric OH stretch)

An example of combination: v=v3 +v2
1/λ = 1/λ3 + 1/λ2 = 0.5089 –> = 1.965 μm

46
Q

• Major spectral regions pertinent to remote sensing (see graph)

A

used in lab

47
Q

The nature and amount of atmospheric absorption depends on (three):

A
  • Absorption spectra of the atmospheric gases
  • Clouds
  • Aerosols
48
Q

*** Atmospheric Windows

A

Regions in the EM spectrum where energy can be fully transmitted (pass through the atmosphere because little absorption)

  • 0.3-0.7 mm –UV and visible light
  • 3-5 mm – emitted thermal energy from Earth
  • 8-11/14 mm –emitted thermal energy from Earth
  • 1 mm-1 m – radar and microwave energy
49
Q

Reviewed from last week

A

transmissivity + reflectivity + absorptivity = 1
albedo
Snell’s law

50
Q

albedo–surface vs spectral

A

tba

51
Q

which in EMR are reflected?

A

visible, near IR

52
Q

Which are emitted?

A

Thermal IR and microwave