()
11
1
22
2
1
1
1
2
sin
sin
1
1
tan
cos
1s
i
n
1/sin
1
/
1
1
0.89.
900nm/600nm
1
d
θ
==
=
=
−
−
λ−
−
64. (a) From the expression for the halfwidth
∆
θ
hw
(given by Eq. 3628) and that for the
resolving power
R
(given by Eq. 3632), we find the product of
∆
hw
and
R
to be
∆
θθ
hw
R
Nd
Nm
m
d
d
d
=
F
H
G
I
K
J
===
λλ
cos
cos
sin
cos
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 Spring '08
 Any
 Physics, Power, Sin, Corresponding angles, Rey Mysterio, Jr., satisfies d sin

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