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})]-1
s' = 22.865 mm
But from the lens the distance is (BFD or back focal distance)
BFD = s' +  δ′ = 22.856 - 0.6849
BFD = 22.180 mm

Solutions 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 mm
In 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
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