64. Let the m = 10 bright fringe on the screen be a distance y from the central maximum. Then from
Fig. 36-8(a)
r
1
− r
2
=
(y + d/2)
2
+ D
2
−
(y
− d/2)
2
+ D
2
= 10λ ,
from which we may solve for y. To the order of (d/D)
2
we find
y = y
0
+
y(y
2
+ d
2
/4)
2D
2
,
where y
0
= 10Dλ/d. Thus, we find the percent error as follows:
y
0
(y
2
0
+ d
2
/4)
2y
0
D
2
=
1
2
10λ
D
2
+
1
8
d
D
2
=
1
2
5.89 µm
2000 µm
2
+
1
8
2.0 mm
40 mm
2
which yields 0.03%.