The nature of electromagnetic radiation part1

Question 1

Electromagnetic radiation in the atmosphere interacts with

Answer 1

gases, aerosol particles, and cloud particles.

Question 2

Extinction and emission are two main types of

Answer 2

the interactions between an electromagnetic radiation field and a medium (e.g., the atmosphere).

Question 3

Extinction is

Answer 3

a process that decreases the radiative intensity, while emission increases it.

Question 4

extinction also means

Answer 4

attenuation

Question 5

Radiation is ………………….. by …………………………………………………………………………..

Answer 5

emitted

all bodies that have a temperature above absolute zero (O K) (often referred to as thermal emission).

Question 6

Extinction is due to

Answer 6

absorption and scattering.

Question 7

Absorption is

Answer 7

a process that removes the radiative energy from an electromagnetic field and transfers it to other forms of energy.

Question 8

Scattering is

Answer 8

a process that does not remove energy from the radiation field, but may redirect it.

Question 9

Scattering can be thought of as ……………………….of …………………………………………….

Answer 9

absorption

radiative energy followed by re- emission back to the electromagnetic field with negligible conversion of energy

Question 10

Scattering can be thought of as absorption of radiative energy followed by re- emission back to the electromagnetic field with negligible conversion of energy. Thus,

Answer 10

scattering can remove radiative energy of a light beam traveling in one direction, but can be a “source” of radiative energy for the light beams traveling in other directions.

Question 11

what is the difference between elastic and inelastic scattering

Answer 11

Elastic scattering is the case when the scattered radiation has the same frequency as that of the incident field. Inelastic (Raman) scattering results in scattered light with a frequency different from that of the incident light.

Question 12

Blackbody is

Answer 12

a body whose absorbs all radiation incident upon it.

Question 13

Properties of blackbody radiation:

Answer 13

• Radiation emitted by a blackbody is isotropic, homogeneous and unpolarized;
• Blackbody radiation at a given wavelength depends only on the temperature;
• Any two blackbodies at the same temperature emit precisely the same radiation;
• A blackbody emits more radiation than any other type of an object at the same temperature;

Question 14

The atmosphere is not strictly in the thermodynamic equilibrium because

Answer 14

its temperature and pressure are functions of position. Therefore, it is usually subdivided into small subsystems each of which is effectively isothermal and isobaric referred to as Local Thermodynamical Equilibrium (LTE).

Question 15

A concept of LTE plays a fundamental role in

Answer 15

atmospheric studies: e.g., the main radiation laws discussed below, which are strictly speaking valid in thermodynamical equilibrium, can be applied to an atmospheric air parcel in LTE.

Question 16

Planck function definition

Answer 16

B_λ(T), gives the intensity (or radiance) emitted by a blackbody having a given temperature.

Question 17

Plank function can be expressed in

Answer 17

wavelength, frequency, or wavenumber domains

Question 18

Plank function can be expressed in wavelength, frequency, or wavenumber domains as

Answer 18

Question 19

what does each symbol mean

Answer 19

Question 20

The relations between B_v~ (T ); B_v (T ) and B_λ(T ) are derived using that

Answer 20

Question 21

Explain the graph

Answer 21

the graph shows plancks radiance “emissivity” and wavelength

when temperature of the object is low wavelength is larger and radiance is low

Question 22

The Stefan-Boltzmann law states that

Answer 22

the radiative flux emitted by a blackbody, per unit surface area of the blackbody, varies as the fourth power of the temperature.

Question 23

Stefan-Boltzmann law formula

Answer 23

Question 24

explain the terms in Stefan-Boltzmann law

Answer 24

Question 25

Wien’s displacement law

Answer 25

states that the wavelength at which the blackbody emission spectrum is most intense varies inversely with the blackbody’s temperature.

Question 26

The Wien’s displacement law states that the wavelength at which the blackbody emission spectrum is most intense varies inversely with the blackbody’s temperature. The constant of proportionality is

Answer 26

Wien’s constant(2897 K µm): λ_m= 2897 / T

Question 27

Explain the terms of weins displacement law

Answer 27

where λm is the wavelength (in micrometers, µm) at which the peak emission intensity occurs, and T is the temperature of the blackbody (in degrees Kelvin, K)

Question 28

Wein's displacement law is simply derived from

Answer 28

dB_λ/dλ= 0

Question 29

Wien's displacement law (easy to remember statement)

Answer 29

The hotter the object the shorter the wavelength of the maximum intensity emitted

Question 30

Kirchhodd's law

Answer 30

states that the emissivity ελ, of a medium is equal to the absorptivity, Αλ, of this medium under thermodynamic equilibrium ελ= Αλ

Question 31

explain the terms of kirchhoff's law

Answer 31

where ελis defined as the ratio of the emitting intensity to the Planck function; Αλis defined as the ratio of the absorbed intensity to the Planck function.

Question 32

**kirchhoff's law** For a blackbody: For a gray body: For a non-blackbody

Answer 32

For a blackbody: ε_λ= Α_λ = 1 For a gray body: ε= Α \< 1 (i.e., no dependency on the wavelength) For a non-blackbody: ε_λ= Α_λ \< 1

Question 33

Kirchhoff’s lawapplies to

Answer 33

gases, liquids and solids if they in TE or LTE

Question 34

In remote sensing applications, one needs to distinguish between the

Answer 34

emissivity of the surface(e.g., various types of lands, ice, ocean etc.) and the emissivity of an atmospheric volume (consisting of gases, aerosols, and/or clouds).

Question 35

Brightness temperature

Answer 35

Tb,is defined as the temperature of a blackbody that emits the same intensity as measured. Brightness temperature is found by inverting the Planck function.

Question 36

where I_λ is the

Answer 36

measured intensity

Question 37

Brightness temperature for a blackbody

Answer 37

brightness temperature = kinetic temperature ( Tb= T)

Question 38

Brightness temperature for natural materials:

Answer 38

T_b⁴ = εT⁴ | (ε is the broadband emissivity)

Question 39

The ocean and land surfaces can modify the atmospheric radiation field by

Answer 39

a) reflecting a portion of the incident radiation back into the atmosphere; b) transmitting some incident radiation; c) absorbing a portion of incident radiation ( Kirchhoff’s law); d) emitting the thermal radiation (Kirchhoff’s law);

Question 40

What does A represent

Answer 40

Incident Radiation

Question 41

What does B represent

Answer 41

Reflected radiation

Question 42

What does C represent

Answer 42

Emitted radiation

Question 43

what does d represent

Answer 43

Transmitted radiation

Question 44

what does e represent

Answer 44

Absorbed radiation

Question 45

Conservation of energy requires that

Answer 45

monochromatic radiation incident upon any surface, I_i, is either reflected, I_r, absorbed, I_a, or transmitted, I_t

Question 46

Conservation of energy requires that monochromatic radiation incident upon any surface, I_i, is either reflected, I_r, absorbed, I_a, or transmitted, I_t, Thus

Answer 46

``` I<sub>i</sub>= I<sub>r</sub> + I<sub>a</sub>+ I<sub>t</sub> 1 = I<sub>r</sub> / I<sub>i</sub> + I<sub>a</sub>/ I<sub>i</sub> + I<sub>t</sub> / I<sub>i</sub> = R + A+ T ```

Question 47

What is T, A and R

Answer 47

where T is the transmission, A is the absorption, and R is the reflection of the surface.

Question 48

In general, T, A, and R are functions of

Answer 48

wavelength: R_λ + A_λ+ T_λ =1

Question 49

Blackbody surfaces (\_\_\_\_\_\_\_\_\_\_\_\_\_\_) and surfaces in LTE (from Kirchhoff’s law):

Answer 49

no reflection

Question 50

Blackbody surfaces (no reflection) and surfaces in LTE (from Kirchhoff’s law):

Answer 50

A_λ = ε_λ

Question 51

Opaque surfaces (\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_-)

Answer 51

no transmission

Question 52

Opaque surfaces (no transmission)

Answer 52

R_λ + A_λ = 1

Question 53

Thus for the opaque surfaces

Answer 53

ε_λ = 1- R_λ

Question 54

In general, emissivity depends on

Answer 54

the direction of emission, surface temperature, wavelength and some physical properties of the surface

Question 55

In the thermal IR (\_\_\_\_\_\_\_\_\_\_\_), nearly all surfaces are

Answer 55

4µm\<λ\< 100µm efficient emitters with the emissivity \> 0.8 and their emissivity does not depend on the direction.

Question 56

In the thermal IR (4µm\<λ\< 100µm), nearly all surfaces are efficient emitters with the emissivity \> 0.8 and their emissivity does not depend on the direction. Therefore,

Answer 56

the intensity emitted from a unit surface area at a given wavelength is I_λ = ε_λ B_λ(T_s)

Question 57

In the shortwave region (\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_), emissivity is

Answer 57

0.1 µm \<λ\< 4 µm negligibly small.

Question 58

In microwave (\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_), emissivity

Answer 58

0.1 cm\<λ\< 100 cm depends on the type and state of the surface.

Question 59

IR window

Answer 59

10 to12 µm

Question 60

Surface: water Emissivity:

Answer 60

0.993-0.998

Question 61

Surface: Ice Emissivity:

Answer 61

0.98

Question 62

Surface: green grass Emissivity:

Answer 62

0.975 - 0.986

Question 63

Surface: sand emissivity:

Answer 63

0.949-0.962

Question 64

Surface: snow emissivity:

Answer 64

0.969-0.997

Question 65

Surface: granite Emissivity:

Answer 65

0.898

Question 66

Many natural surfaces have \_\_\_\_\_\_\_\_\_\_emissivity in the IR window and hence \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_--

Answer 66

high low (negligible) reflectivity in this spectral region.