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52.
(a) From the expression for the halfwidth ∆
θ
hw
(given by Eq. 3725) and that for the resolving power
R
(given by Eq. 3729), we ±nd the product of ∆
θ
hw
and
R
to be
∆
θ
hw
R
=
µ
λ
Nd
cos
θ
¶
Nm
=
mλ
d
cos
θ
=
d
sin
θ
d
cos
θ
=tan
θ,
where we used
mλ
=
d
sin
θ
(see Eq. 3722).
(b) For ±rst order
m
= 1, so the corresponding angle
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Power

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