Quote:
Originally Posted by Ryderscope
I am going to make a couple of assumptions here. The first one is that the UV/IR cutoff filter will affect the length of the optical path in the same way as the filters that we use for monochrome imaging. So if this assumption is correct then, based on traditional rules of thumb, the length of the optical path will be lengthened by an amount equal to 1/3 of the thickness of the glass in the UV/IR cutoff filter.
I have been unable to find any specifications for the thickness of the glass in the UV/IR cutoff filters so the second assumption that I am going to make is that they are 1mm thick. The increase in the length of the optical path due to the filter then is 1/3 of 1mm or 0.33mm. If anyone has any detailed knowledge of this please share.
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Hi R,
As an exercise of the grey matter, alegbra, trig, optics/physics etc... I decided to flesh out what I thought was the basis for the 1/3rd thickness of the filter rule of thumb....
It looks like it's quite a valid supposition if you take the refractive indices involved as ~1 for air (1.0003) and ~1.5 for the glass (this could be anywhere from 1.4 to 1.7 depending on glass used). Anyway if you're interested in the derivation of the difference in focal length of an optic WITH and WITHOUT an intermediary optical layer (glass) just completed, then see the attached.
Long story short .... SEE THE
RED BOX. If you substitute typical values for refractive indices of air and glass of 1 and 1.5 respectively you will note that the focal point shifts backwards from POINT
A in the diagram to POINT
B in the diagram by a distance of 1/3rd the filter thickness, when the filter is included in the optical path. Given by the derived expression ...
b = x [1 -(n1/n2)]
where
b = distance focal point moves with and without filter (see diagram)
x = glass / filter thickness
n1 = refractive index of medium before filter (i.e: that of the air)
n2 = refractive index of the filter medium (i.e: that of the glass)
Best
JA