Optics Tutorial 2 HW solutions

1.An equi-convex fused Silica lens has a 90 radius of curvature. Compute the focal length. 

Solution: Use the thin lens equation to calculate focal length: Power = φ = 1/F ≈ (n-1) ((1/R1) –(1/R2)) = (n-1)(C1-C2)

Fused silica has a visible index of 1.4585 [Source]

Caveat: the back radius is negative. If one uses a positive radius of curvature then you are calculating the focal length for a meniscus lens.

F ≈ [(n-1) * ((1/R1) –(1/R2))]-1 
 [(1.4585-1) * ((1/90) –(1/{-90}))]-1 = [(1.4585-1) * ((1/90) +(1/90))]-1 = [(1.4585-1) * (2/90)]-1 = 90 / 2 / (1.4585-1) = 45 / 0.4585
 98.15 

Holy Frioles! that algebra is a mess!

2.A lens has a front convex radius of 200 mm and a rear concave radius of 1000 mm. What is the shape factor? Flip lens and re-compute the shape factor.

Solution: in the first incarnation both radii of curvature are positive, since both center of curvatures are to the right.
S = (RFront + RBack) / (RBack – RFront) =  (200 + 1000) / (1000 – 200)
S = 1.5, Positive Meniscus

The second incarnation or a flipped lens, both radii of curvature are negative, since both center of curvatures are to the left.
S = (RFront + RBack) / (RBack – RFront) =  (-1000 + (-200)) / (-200 – (-1000))
S = -1.5, Positive Meniscus

Conclusion: Flipping a lens changes the sign of the shape factor. Why do we care about this parameter? It will become more apparent when we begin to study spherical aberration. For a distant star imaged by this lens each lens orientation will have a different amount of spherical aberration and therefore different spot sizes. Aside: "... a distant star..." is referred to as an "infinite conjugate", but I hesitated using overly verbose optical engineering nomenclature:)
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