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AMP EXAM – 22-01-2021 Proff. Annoni, Demir, Strano Question 1 You need to perform a keyhole welding of 3 mm thickness stainless steel plates. You can use a fiber laser source with a power of 6 kW. The fiber core diameter is 400 μm and the numerical aperture is NA=0.24. Your laser head can be equipped with two different collimation lenses and two different focusing lenses. All the lenses are characterized by a lens diameter of 60 mm, the focal lengths are: -Focal length of first collimation lens, fcol1=80 mm; -Focal length of second collimation lens, fcol2=120 mm; -Focal length of first focusing lens, ffoc1=150 mm; -Focal length of second focusing lens, ffoc2=200 mm. Q: It is asked to choose the optical configuration able to weld correctly the stainless steel plates. Assumption: the position of waist diameter is in the middle of thickness plates. The diameter of the collimation lens must be at least 20% bigger than the collimated beam one (dc). Sol The first step is determine de half-angle of divergence at the exit of the fiber by the information on Numerical Aperture: 𝜃"=sin'(𝑁𝐴=0.24 𝑟𝑎𝑑 ≈14° Using trigonometry it is valid the relationship: 𝜃"=𝑑62𝑓689 From which as a function of different collimation lens it is possible determine the collimated beam diameter (i.e. the value of dc must be check with the restriction on collimation lens diameter à The diameter of the collimation lens must be at least 20% bigger than the collimated beam one (dc)) With fcol1: 𝑑6(=38.8 mm < 0.2 Dlens (YES!) With fcol2: 𝑑6:= 58.2 mm > 0.2 Dlens (NOT FEASIBLE!) Considering the choice of fcol1 it is possible determine two different waist diameter as a function of the two option of focusing lens. With ffoc1 𝑑;(=𝑑""<=>?">=@?=0.75 𝑚𝑚 With ffoc2 𝑑;:=𝑑""<=>D">=@?=1 𝑚𝑚 Now considering the application…