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59. (a) We note that
d
= (76
×
10
6
nm)/40000 = 1900 nm. For the first order maxima
λ
=
d
sin
θ
, which leads to
=
F
H
G
I
K
J
=
F
H
G
I
K
J
=°
−−
sin
.
11
589
1900
18
λ
d
nm
Now, substituting
m = d
sin
/
λ
into Eq. 3630 leads to
D
= tan
/
λ
= tan 18°/589 nm = 5.5
×
10
–4
rad/nm = 0.032°/nm.
(b) For
m
= 1, the resolving power is
R = Nm
= 40000
m
= 40000 = 4.0 × 10
4
.
(c) For
m
= 2 we have
= 38°, and the corresponding value of dispersion is 0.076°/nm.
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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

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