### Optics Tutorial 8 HW solutions

 A lens with and index of 3.0 and 10 mm thick has a front convex radius of 200 mm and a back convex radius of 50 mm.  1.What is the lens focal length? Solution: 1/F = (n-1) {(1/R1) –(1/R2)+(n-1)/n*t/R1/R2} F = [ (n-1) {(1/R1) –(1/R2)+(n-1)/n*t/R1/R2}]-1 F = [ (3-1) {(1/200) –(1/{-50})+(3-1)/3*10/200/(-50)}]-1 F = 20.5479 mm 2.What are the vertex to principle plane distances? Solution: δ = t* F/R2 * (n-1)/n = 10 * 20.5479/50 * (3-1)/3 (Screwed up the sign of this equation in the lecture) δ = +2.740 mm δ′ = -t * F/R1 * (n-1)/n = -10 * 20.5479 / 200 * (3-1)/3 δ′ = -0.6849 mm 3.What is the hiatus thickness Solution: T - δ  + δ′ = 10 - 2.740 - (-0.6849) Hiatus = 6.4750 4.For a 200 mm object distance (object to first principle plane), what is the image distance from the last vertex? Solution: Use the lens makers equation 1/s' = (1/F) + (1/s) But s = 200 + δ = 202.749 s' = [(1/F) + (1/s)]-1 = [(1/20.5479) + (1/{-202.749})]-1s' = 22.865 mmBut from the lens the distance is (BFD or back focal distance)BFD = s' +  δ′ = 22.856 - 0.6849BFD = 22.180 mmSolutions 1 through 4 are verified via a Zemax model attached below 5.If the object space is water, where are the nodal points? Solution:Nodal Shift = (n' - n) * F = (1-1.3333) * 20.5479 mm (reference for index of water)Nodal shifts = -6.842 mmIn this case the front nodal point is outside the lens.
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Optics Tutorial 8, HW solutions.SES
(79k)
Scott Sparrold,
Mar 7, 2013, 2:03 PM
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Optics Tutorial 8, HW solutions.ZMX
(2k)
Scott Sparrold,
Mar 7, 2013, 2:03 PM