44. At the point on the screen where we find the inner edge of the hole, we have tan θ = 5.0 cm/30 cm,
which gives θ = 9.46
◦
. We note that d for the grating is equal to 1.0 mm/350 = 1.0
× 10
6
nm/350. From
mλ = d sin θ, we find
m =
d sin θ
λ
=
1.0
×10
6
nm
350
(0.1644)
λ
=
470 nm
λ
.
Since for white light λ > 400 nm, the only integer m allowed here is m = 1. Thus, at one edge of the
hole, λ = 470 nm. However, at the other edge, we have tan θ
= 6.0 cm/30 cm, which gives θ
= 11.31
◦
.
This leads to
λ
= d sin θ
=
1.0
× 10
6
nm
350
sin 11.31
◦
= 560 nm .
Consequently, the range of wavelength is from 470 to 560 nm.