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 .