p37 021

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21. Eq. 37-14 gives θ

R

= 1.22λ/d, where in our case θ

R

≈ D/L, with D = 60 µm being the size of the object

your eyes must resolve, and L being the maximum viewing distance in question. If d = 3.00 mm =
3000 µm is the diameter of your pupil, then

L =

Dd

1.22λ

=

(60 µm)(3000 µm)

1.22(0.55 µm)

= 2.7

× 10

5

µm = 27 cm .


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