1. Radiometric Concepts Remote Sensing ERAU Dr. Darrel Smith September 30, 2008 Remote Sensing ERAU Dr. Darrel Smith September 30, 2008

2. Outline  Radiometric Terms  Blackbody radiation  Inverse-Square Law for Irradiance  Lambertian Surfaces  Radiometric Terms  Blackbody radiation  Inverse-Square Law for Irradiance  Lambertian Surfaces

3. Radiometric Terms  Irradiance (E) --flux per unit area onto a surface  Radiant Exitance (M)--flux per unit area away from the surface.  Irradiance (E) --flux per unit area onto a surface  Radiant Exitance (M)--flux per unit area away from the surface.

4. Radiometric Terms  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.  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.

5. Blackbody Radiation  Blackbody spectrum -- Planck (1901) Emissivity (  ) -- ratio of the spectral exitance to the exitance from a blackbody at the same temperature.  0 <  < 1  Blackbody spectrum -- Planck (1901) Emissivity (  ) -- ratio of the spectral exitance to the exitance from a blackbody at the same temperature.  0 <  < 1

6. Blackbody Radiation   = constant describes “gray bodies”   = varies with describes “selective radiators”   = constant describes “gray bodies”   = varies with describes “selective radiators”

7. Transmission, Reflection, Absorption  Transmission Reflection Absorption Conservation of Energy Note: Kirchoff’s law  = 

8. Properties of Blackbody Radiation  Calculate the “total exitance” from a blackbody radiator. where  = 5.67  10 -8 W/(m 2  K 4 )  Peak of the blackbody exitance (Wien’s Law) where A = 2898  m  K  Calculate the “total exitance” from a blackbody radiator. where  = 5.67  10 -8 W/(m 2  K 4 )  Peak of the blackbody exitance (Wien’s Law) where A = 2898  m  K Note: 10  m window at T = 300K

9. Homework Problem  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.

10. Stealth Technologies How does stealth technology work? Reduce the RCS (Radar Cross Section) 1.Choice of angles 2.Materials that act like a blackbody

11. Lambertian Surfaces  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.  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.

12. Lambertian Surfaces  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:

13. Lambertian Surface 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. 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.