Visibility on Venus

Took the mg/m³ values and tried to figure out how they affected visibility.  I came up with a solid source that stated 0.1 mg/m³ was 7.5 km in dry air (Venus has 0% RH).


The plot looks linear but if we extrapolate to 0.001 mg/m³ it would be 750 km in a layer that is purportedly “haze” which is, by definition, 5 km or less. They I ran across this graph that used the Extinction Coefficient instead of density.


Which led me to Koschmieder’s Relationship: If k(e) is σ then V = 3/σ or 3 km at 60 km… seems a little high!  Is that because water droplets are so much larger?  If k(e) is in km then there is no further conversion, right?

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