1 / 13

# Radiometric Concepts - PowerPoint PPT Presentation

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Remote Sensing

ERAU

Dr. Darrel Smith

September 30, 2008

• Lambertian Surfaces

• Irradiance (E) --flux per unit area onto a surface

• Radiant Exitance (M)--flux per unit area away from the surface.

• Radiant Intensity (I) --describes the angular distribution of the flux from a point source.

• The “radiance” (L)--provides both angular and spatial information about the flux from a point source.

• Blackbody spectrum -- Planck (1901)Emissivity () -- ratio of the spectral exitance to the exitance from a blackbody at the same temperature.

• 0 <  < 1

•  = constant describes “gray bodies”

•  = varies with  describes “selective radiators”

• Transmission ReflectionAbsorptionConservation of Energy

Note: Kirchoff’s law  = 

• Calculate the “total exitance” from a blackbody radiator.where  = 5.6710-8 W/(m2K4)

• Peak of the blackbody exitance (Wien’s Law)where A = 2898 mK

Note: 10 m window at T = 300K

• What fraction of the spectral radiant exitance M is in the visible spectrum (400 nm 700 nm) for the sun?Assume a temperature of 5800 K.

• How does stealth technology work?

• Reduce the RCS (Radar Cross Section)

• Choice of angles

• Materials that act like a blackbody

• How is the energy leaving a surface angularly distributed into the hemisphere above the surface?

• A lambertian surface has the following property:

• How is the radiance angularly distributed from a Lambertian surface? Note: the visual response is proportional to the radiance.

• The radiance along the normal from a Lambertian surface will be:The radiance into any direction  from the normal is:Combining the previous 3 equations, we obtain:

• Note: Since perceived brightness is proportional to the radiance in the visible region, this means that a Lambertian surface would look the same from all direction.

• While we cannot assume that all surfaces are Lambertian, it is a good starting point for discussion of less well-behaved surfaces.