White Light Interferometry (WLI) Testing Technique - white light interferometry
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by VE Leshchenko · 2015 · Cited by 59 — For instance to achieve 95% efficiency (ηI) under D0 = 10 mm σθ should be less than 8.4 µrad. Whereas for combining of 20 mm beams with 95% efficiency σθ should ...
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First, the NA is defined in free space, not inside the fiber. So, NA = 0.22 means that is the NA in air. Therefore, the angle is $sin(\theta) = 0.22 \rightarrow \theta = 12.7 deg$. [no need to divide by the index of refraction]
0.3mm. 0.32mm. 80. 0.35mm. 79. 0.38mm. 1/64. 0.4mm. 78. 0.42mm. 0.45mm. 77. 0.48mm ... 3.5mm. 28. 9/64. 3.6mm. 27. 3.7mm. 26. 25. 3.8mm. 24. 3.9mm. 23. 5/32. 22.
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Finally, please appreciate that collimation is identical to imaging at infinity. What this means in practice is two-fold
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Next, NA refers to the half angle, not the full angle (see: http://en.wikipedia.org/wiki/Numerical_aperture ). So: $ tan(\theta) = \frac{D/2}{f} = \frac{38 mm/2}{f}$ which gives you $f = \frac{17.5 mm}{tan(\theta)} = 77.6 mm$ .
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Principal plane are those plane where normal stress are maximum or minimum or you can say shear stress is zero and principal stress are the max ...
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Apr 1, 2024 — 230) a comparison of resolution under different conditions is given using K: K = x⋅NA/λ where x is the interval of a grating, or separation of ...
The quality of the collimated beam may not be fantastic. And it depends on how the light is launched into the fiber. If the incident light has a narrow angular dispersion, and the fiber is short and/or of very good quality with few bends, then the light coming out will have a low angular dispersion. But if the incident light is converging at an angle close to the NA of the fiber, then your analysis should be ok.
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$$ n\sin(\theta)=NA \\ \theta = \arcsin(\frac{.22}{1.4914}) \\ \frac{.5 \cdot 38 [mm]}{ \tan(\theta)}=\left \{ \text{focal length} \right \} $$