54. We are given the “number of lines per millimeter” (which is a common way to
express 1/
d
for diffraction gratings); thus,
1
d
=
160
lines/mm
⇒
d
= 6.25
×
10
−
6
m .
(a) We solve Eq. 3625 for
θ
with various values of
m
and
λ.
We show here the
m =
2
and
λ
= 460 nm calculation:
()
9
11
1
6
10 m)
sin
sin
sin
0.1472
8.46 .
6.25 10 m
m
d
λ
−
−−
−
−
⎛⎞
2(460×
==
=
=
°
⎜⎟
×
⎝⎠
Similarly, we get 11.81° for
m =
2 and
λ
= 640 nm, 12.75° for
m =
3 and
λ
= 460 nm,
and 17.89° for
m =
3 and
λ
= 640 nm.
The first indication of overlap occurs when we
compute the angle for
m =
4 and
λ
= 460 nm; the result is 17.12° which clearly shows
overlap with the largewavelength portion of the
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Diffraction

Click to edit the document details