2 Electromagnetic Energy

Question 1

Basic requirement for Remote Sensing

Answer 1

Source of energy in the form of EM radiation

EM radiation emitted by the source (usually the sun) -> propagates through a material/vacuum -> Interacts with the Earth’s atmosphere -> interacts with the Earth’s surface -> Interacts with the Earth’s atmosphere again -> Reaches the sensor and being recorded by the detectors

Question 2

EM radiation

Answer 2

a form of energy and has wave-like and particle-like properties.

can be modeled by Classic physics and modern quantum theory
– Classical physics (Electromagnetic wave theory)
• The flow of energy through a medium in the form of electric and magnetic fields
– Modern quantum theory (Planck’s quantum theory)
• Flow of photons, as small particles, at the speed of light through space

Question 3

EM waves

Answer 3

• EM waves are created by the vibration of electric charges.
• EM waves are in the form of mutually linked electric (E) and magnetic (M) fields.
• Electric and magnetic fields change in magnitude in a direction perpendicular to the traveling direction of the radiation.
• The magnetic field is oriented at right angles to the electrical field.

Question 4

Important characteristics of EM waves

Answer 4

# Frequency (f)
• How often a wave passes a given point per second
• Measured in hertz (Hz) or its factors such as MHz (106 Hz) or GHz (109 Hz).
• 1 cycle per second = 1 Hz

Wavelength (lamda)
• The length of one cycle (distance between successive crests or troughs)
• Usually measured in m or its factors such as nm (10-9 m), μm (10-6metres) or cm (10-2 meters)

# Speed (c)
• In vacuum, EM waves travel at the speed of light (c)

*note: lamda= c/f. Frequency and wavelength are inversely proportional

Question 5

Planck’s quantum theory

Answer 5

• Planck’s quantum theory models some aspects that could not be rationalized by EM wave theory.
• Electromagnetic radiation exists in discrete quanta of energy, called photons.
• Photons are packets of energy that travel at the speed of light.
• Energy carried by a single photon is proportional to its frequency and inversely proportional to its wavelength

E = hf = hc/lamda
where
E = enrgy 
h = Planck's constant = 6.26 x 10^-34
f = wave frequency 
lamda = wavelength
c = speed of light = 3 x 10^8 m/s

• EM radiation with shorter wavelength have higher level of quantum energy
– Gamma- and X-ray and Ultraviolet are dangerous to human body.
– longer wavelength (lower energy) like Microwaves or radio waves are not dangerous.

• Implication to remote sensing
– It is more difficult to detect the radiant energy at longer wavelength than shorter wavelength

Question 6

Spectral density of EM radiation (1/3)

Answer 6

• Planck’s law describes the spectral density of EM radiation emitted by a black body
• A black body
– Absorbs all incident radiations
– Emits black body radiation at all frequencies

Question 7

Spectral density of EM radiation (2/3)

Answer 7

• Planck’s law describes the electromagnetic radiation a black body emits at a certain wavelength as a function of its temperature

• Kelvin and Celsius have the same magnitude
– 1 K = 1 °C
– [K] = [°C] + 273.15

Question 8

Spectral density of EM radiation (3/3)

Answer 8

• Materials at different temperature emit different magnitudes of EM radiations

• The sun
– Temperature of 5778 K
– Mostly emits in visible, infrared, and UV

• Our planet
– Average temperature of 288 K
– Emits mostly infrared radiation

Question 9

Total EM radiation

Answer 9

• Stefan-Boltzmann’s law describes the total electromagnetic radiation emitted across all wavelengths by a black body

M=§T^4
where 
M = total radiant exitance [Wm^2]
T = absolute temperature [K]
§ = Stefan-Boltzman constant [5.67 x 10^-8 W m^-2 K^-4]

• The total energy is directly proportional to the fourth power of the temperature

Question 10

Maximum Spectral radiation

Answer 10

# The Wien's displacement law 
lamda max = A/T 
where 
lamda max = wavelength of maximum spectral radiant exitance [miu m]
A = Wien's constant [2897.8 mium.K]
T = absolute temperature [K]

– Increasing temperature maximum radiation shifts towards shorter wavelengths
– The temperature of the Sun is 5778 K corresponding to a peak emission at 502 nm

Question 11

the implication of maximum spectral radiation in RS

Answer 11

– To measure human body at average temperature of 310 kelvin and a peak at 10 μm, we need a sensor at the range of 7-14 μm

– To measure earth surface at average temperature of 300 k, we should look at 8-14 μm

– To measure forest fire at 800 k we should look at 3-5 μm.

Question 12

Radiation of gray body

Answer 12

• Real world materials are not ideal black bodies
• Gray bodies emit only a fraction of radiation emitted by black body.
• Effectiveness of an object in emitting EM radiation is defined by Emissivity

Emisivity= radiant exitance of an object at a given temp / radiant exitance of a blackbody at the same temp

–Emissivity values can range from 0 to 1 (dimensionless)
– A black body has an emissivity of 1
– A white body has an emissivity of 0
– A gray body has an emissivity between 0 and 1

Question 13

Selective Radiators

Answer 13

• For many materials, the emissivity depends on the wavelength
– Such material is called selective radiator

• Most natural objects are selective radiators
• Emissivity depends on surface TEMPERATURE; WAVELENGTH; PHYSICAL PROPERTIES of the surface
• Objects with the same Temperature can have significantly different radiations.

Question 14

Interaction between EM radiation and materials

Answer 14

# Absorption
# Transmission
# Reflection
# Scattering

Question 15

Interaction of EM energy with the atmosphere
1. ABSORPTION

2. Transmission
3. Scattering
4. Reflection

Answer 15

– Some parts of EM radiation are absorbed by different gasses in the atmosphere

– Oxygen (O2), Ozone (O3), water vapor (H2O), and carbon dioxide (CO2) are the major absorbers.

Question 16

Interaction of EM energy with the atmosphere

1. Absorption
2. TRANSMISSION
3. Scattering
4. Reflection

Answer 16

– Atmosphere transmits parts of EM radiation

Question 17

Interaction of EM energy with the atmosphere

1. Absorption
2. Transmission
3. SCATTERING
4. Reflection

Answer 17

– Interaction with large particles in the atmosphere may cause EM radiation to change its direction
– The EM energy is absorbed and immediately emitted
– Direction changes, but WAVELENGTH and ENERGY remain intact

Question 18

Interaction of EM energy with the atmosphere

1. Absorption
2. Transmission
3. Scattering
4. REFLECTION

Answer 18

•# Types of reflection
– SPECULARreflection
• Mirror-like reflection
• Occurs when the surface is smooth.
• A major part of energy is reflected.
– DIFFUSE reflection
• Occurs when the surface is rough with respect to incident wavelength
• Energy is reflected in all directions.

In reality, the surfaces are a combination of both types.

Question 19

Scattering depends on

Answer 19

– Wavelength
– Amount of particles in the atmosphere
– Travelling distance of EM radiation in the atmosphere

Question 20

Three types of scattering based on particle size

Answer 20

– Rayleigh scattering
– Mie scattering
– Non-selective scattering

Question 21

Rayleigh scattering

Answer 21

When the particle size is smaller than the EM wavelength
– For example:
• Nitrogen and Oxygen molecules scatter visible spectrum
• Raindrops scatter Microwaves

Strongly dependent on wavelength
– The intensity of scattering is inversely proportional to the fourth power of the wavelength I ~ (1/lamda^4)
– Shorter wavelength EM radiations are scattered more than longer wavelengths
- Blue light is scattered more than red light
– Rayleigh scattering makes blue part of spectrum less useful in remote sensing.

Question 22

Mie scattering

Answer 22

# When the particle size has a similar size to the EM wavelength
– For example: Aerosol scatters visible spectrum

Aerosol is mixture of gases, water vapor, and dust.

Mie scattering has very little dependence on wavelength

In longer wavelengths, Mie scattering is higher than Rayleigh scattering.

Question 23

Non-selective scattering

Answer 23

#When the particle size is significantly larger than wavelength
– For example: Water droplets scatter visible spectrum

All radiation is evenly scattered through the visible and infrared spectrum

Clouds scatter all wavelengths of EM radiation, therefore they appear white.

Non-selective scattering makes areas prone to cloud less useful in visible and infrared parts of spectrum

Question 24

Why the sky is red at sunset?

Answer 24

Because of Raleigh scattering, when the sun is on low position during sunset, the EM radiation from the sun has to travel farther to reach our eyes. The red part of EM radiation from the sun can travel this distance without being too much scattered while the blue part had scattered. That’s how sky appears red in at sunset.

Question 25

Why clouds are white?

Answer 25

The EM radiation is scattered in many wavelengths by the water droplet in the clouds. As all wavelengths are scattered equally and mixed, clouds appear white to us.

Question 26

Why clouds are white?

Answer 26

The EM radiation is scattered in many wavelengths by the water droplet in the clouds. As all wavelengths are scattered equally and mixed, clouds appear white to us (non-selective scattering).

Question 27

The implication of interaction between EM energy and atmosphere in optical remote sensing

Answer 27

– Absorption • Absorption reduces the electromagnetic radiance that reaches earth surface within the absorption bands. • The reflected radiance is also attenuated after passing through the atmosphere. • Because the absorption is wavelength-dependent, the apparent spectral signature of the target is altered. – Scattering • Scattering degrades remote sensing images. • Solar radiation scattered by the atmosphere towards the sensor before reaching the earth causes a hazy appearance of the image. • Rayleigh scattering causes a strong hazy effect in shorter wavelengths i.e. blue. • EM radiance from targets neighboring targets can be scattered into sensor’s field of view • Reflected radiation can be scattered making the target blur.

Question 28

Interaction of EM energy with earth’s surface

Answer 28

• Absorption • Transmission – e.g. through water • Reflection – In many types of remote sensing, solar energy bounced off the earth’s surface is measured by the sensor. • The proportion of reflected, absorbed, and transmitted

Question 29

Spectral Signature

Answer 29

* Variation of reflectance (or emittance) of a material as a function of wavelengths * Reflectance measurement can be carried out in the laboratory or in the field using a field spectrometer.

Question 30

Reflectance characteristics of materials -Vegetation

Answer 30

– Reflectance at a given wavelength depends on the leaf properties. – The reflectance in blue and red components are relatively low because they are absorbed by the plant for photosynthesis. – In Near Infrared (NIR), the reflectance is highest, but the amount depends on the leaf development and cell structure. – In Shortwave Infrared (SWIR), the reflectance is controlled by the free water in the leaf tissue. • More free water results in less reflectance – At the harvest time, no photosynthesis is done. This results in higher reflectance in red part. – The leaves are drier resulting in higher reflectance at SWIR – The reflectance at NIR may increase

Question 31

Reflectance characteristics of materials -Water

Answer 31

– Comparing to vegetation, water has a lower reflectance. – At most 10% of the incident energy is reflected. – Most of reflectance is in visible and a little in NIR range. – Reflectance change if there is sediment or plants in the water.