Wave Optics
329
24.58
The wavelength is
340 m s
0.170 m
2 000 Hz
sound
v
f
λ
=
=
=
sin
d
m
Maxima occur where
θ
λ
=
, or
(
)
1
sin
m
d
θ
λ
−
=
for
0, 1, 2,
m
=
…
Since
,
0.350 m
d
=
0.486
d
λ
=
which gives
(
)
1
sin
0.486
m
θ
−
=
For
, this yields
0,1, and 2
=
m
max
3
m
≥
ima at 0
, 29.1
, and 76.
°
°
sin
3
°
1
No solutions exist for
since that would imply
θ
>
Minima occur where
(
)
sin
d
m
1 2
θ
λ
=
+
or
(
)
1
sin
2
m
1
2
d
λ
θ
−
+
=
=
for
m
With
0, 1, 2,
…
0.486
d
λ
=
, this becomes
(
)(
)
1
0.243
m
=
+
1
sin
−
2
θ
For
, we find
0 and 1
=
m
minima
2
m
≥
at 14.1
and 46.8
°
°
sin
1
No solutions exist for
since that would imply
θ
>
24.59
The source and its image, located 1.00 cm below the mirror, act as a pair of coherent
sources. This situation may be treated as doubleslit interference, with the slits separated
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 Spring '09
 Physics, Work, Wavelength, Sin, WAVE OPTICS

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