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⇤ ← Revision 1 as of 2015-07-24 01:31:35
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| where $ c $ is the speed of light. Defining $\delta\lambda~=~\lambda_1-\lambda_2 $ and center wavelength $ \lambda = \sqrt{\lambda_1\lambda_2}$, this simplifies to: | where $ c $ is the speed of light. Defining $\Delta\lambda~=~\lambda_1-\lambda_2 $ and center wavelength $ \lambda = \sqrt{\lambda_1\lambda_2}$, this simplifies to: |
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| $ Jy ~=~ 1e26*(P/A)*c~\delta\lambda/\lambda^2 $ | $ Jy ~=~ 1e26*(P/A)*c~\Delta\lambda/\lambda^2 $ |
Infrared Telescopes
Jansky Units and Infrared Radio Telescopes
1 Jansky is 1e-26 W/m2-Hz. Hz implies bandwidth. The bandwidth of a radio telescope is obvious - the bandwidth in Hertz, probably after a chilled reflector into a chilled low noise amplifier, then through a filter. A wider filter picks up more power, and more thermal noise. So the power in Janksy units for a source is Jy = 1e26 * Power~Received / ( Area * ( f_2 - f_1 ) with power in watts, Area of the receiving surface in square meters, and frequencies in Hertz.
Infrared telescopes have much wider bandwidth, and infrared is typically characterized with wavelength (usually micrometers) rather than Hertz. We can recast the formula as
Jy~=~1e26*(P/A)*c*(1/\lambda_2-1/\lambda_1)~=~1e26*(P/A)*c(\lambda_1-\lambda_2)/(\lambda_1\lambda_2)
where c is the speed of light. Defining \Delta\lambda~=~\lambda_1-\lambda_2 and center wavelength \lambda = \sqrt{\lambda_1\lambda_2}, this simplifies to:
Jy ~=~ 1e26*(P/A)*c~\Delta\lambda/\lambda^2