85.

(a) Letting d sin θ = mλ, we solve for λ:

λ =

d sin θ

m

=

(1.0 mm/200)(sin 30

◦

)

m

=

2500 nm

m

where m = 1, 2, 3

· · ·. In the visible light range m can assume the following values: m

1

= 4, m

2

= 5

and m

3

= 6. The corresponding wavelengths are λ

1

= 2500 nm/4 = 625 nm, λ

2

= 2500 nm/5 =

500 nm, and λ

3

= 2500 nm/6 = 416 nm.

(b) The colors are orange (for λ

1

= 625 nm), blue-green (for λ

2

= 500 nm), and violet (for λ

3

= 416 nm).