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81. (a) We express all lengths in mm, and since 1
/d
= 180, we write Eq. 3722 as
θ
= sin
−
1
µ
1
d
mλ
¶
= sin
−
1
(180)(2)
λ
where
λ
1
=4
×
10
−
4
and
λ
2
=5
×
10
−
4
(in mm). Thus, ∆
θ
=
θ
2
−
θ
1
=2
.
1
◦
.
(b) Use of Eq. 3722 for each wavelength leads to the condition
m
1
λ
1
=
m
2
λ
2
for which the smallest possible choices are
m
1
=5and
m
2
= 4. Returning to Eq. 3722, then, we
Fnd
θ
= sin
−
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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

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