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.