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%.