50.
(a) From d sin θ = mλ we find
d =
mλ
avg
sin θ
=
3(589.3 nm)
sin 10
◦
= 1.0
× 10
4
nm = 10 µm .
(b) The total width of the ruling is
L = N d =
R
m
d =
λ
avg
d
m∆λ
=
(589.3 nm)(10 µm)
3(589.59 nm
− 589.00 nm)
= 3.3
× 10
3
µm = 3.3 mm .