›› rand(foo); Adam Geboff

24Mar/100

F/# as an angle

i have recently come across an issue with designing an optical system to interface with an Oriel MS257 monochrometer and found people referring to the divergence of the instrument by the F/# listed in the MS257's spec sheet - an F/3.9 at the input slit.  This didn't make sense to me as F/# is defined as the ratio of a lens's focal length to its aperture.
After some digging, it turns out that optics people use F/# and numerical aperture (NA) as ways of expressing a cone angle which, for the rest of this post, i will refer to as θ.  If we define the half angle using tangent, θ/2 is defined by the ratio of its opposite length to its adjacent length which easily correlate to half a lens' aperture and its focal length respectively. Using basic trigonometry we get:

\begin{matrix}\frac{\theta}{2}&=&\arctan\left(\frac{d/2}{fl}\right)\\\theta&=&2\cdot\arctan\left(\frac{d/2}{fl}\right)\end{matrix}

where d is the length of the aperture and fl is the focal length.  We can then massage this equation such that it is in terms of F/#:

NOTE:  F/\#=\frac{fl}{d}

\begin{matrix}\theta&=&2\cdot\arctan\left(\frac{d/2}{fl}\right)\\&=&2\cdot\arctan\left(\frac{1}{2}\cdot\frac{d}{fl}\right)\\&=&2\cdot\arctan\left(\frac{1}{2}\cdot F/\#^{-1}\right)\end{matrix}

So drawing from the MS257 example where we have F/3.9 at the input, that translates to an input angle of 14.61°, and any source that we input into the monochrometer must be focusing at the input slit at an angle of 14.61°.

If we want to relate F/# to NA, we must start with the definition of numerical aperture which is defined by the equation:

NA=n\cdot\sin\left(\phi\right)

where n is in index of refraction and in this situation it is ≈1, and φ is the half cone angle which we have been referring to as θ/2.  We defined θ above and when we combine these two equations we get:

\begin{matrix}NA&=&n\cdot\sin\left(\phi\right)\\&=&1\cdot\sin\left(\frac{\theta}{2}\right)\\&=&\sin\left(\frac{2\cdot\arctan\left(\frac{1}{2}\cdot F/\#^{-1}\right)}{2}\right)\\&=&\sin\left(\arctan\left(\frac{1}{2}\cdot F/\#^{-1}\right)\right)\end{matrix}

To again use our MS257 example of F/3.9, the input slit of the monochrometer has an NA of 0.127.

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