84. (a) We require that sin
θ
=
mλ
1
,
2
/d
≤
sin 30
◦
,where
m
=1
,
2and
λ
1
= 500 nm. This gives
d
≥
2
λ
s
sin 30
◦
=
2(600 nm)
sin 30
◦
= 2400 nm
.
For a grating of given total width
L
we have
N
=
L/d
∝
d
−
1
, so we need to minimize
d
to maximize
R
=
mN
∝
d
−
1
.Thu
sw
echoo
se
d
= 2400 nm.
(b) Let the thirdorder maximum for
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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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