Diffraction
3619
36.72.
I
DENTIFY
:
Apply
sin
1.22
D
λ
θ
=
.
S
ET
U
P
:
θ
is small, so
sin
x
R
θ
Δ
≈
, where
x
Δ
is the size of the details and
R
is the distance to the earth.
15
1 ly
9.41
10
m
=
×
.
E
XECUTE
:
(a)
6
5
17
5
(6.00
10 m)(2.50
10 m)
1.23
10
m
13.1ly
1.22
(1.22)(1.0
10
m)
D x
R
λ
−
Δ
×
×
=
=
=
×
=
×
(b)
5
15
8
1.22
(1.22)(1.0
10
m)(4.22 ly)(9.41 10
m ly)
4.84
10 km
1.0 m
R
x
D
λ
−
×
×
Δ
=
=
=
×
.
This is about 10,000 times the
diameter of the earth! Not enough resolution to see an earthlike planet!
x
Δ
is about 3 times the distance from the
earth to the sun.
(c)
5
15
6
6
(1.22)(1.0
10
m)(59 ly)(9.41 10
m
ly)
1.13
10 m
1130 km.
6.00
10 m
x
−
×
×
Δ
=
=
×
=
×
3
5
planet
1130 km
8.19
10
;
1.38
10 km
x
x
D
−
Δ
=
=
×
Δ
×
is small compared to the size of the planet.
E
VALUATE
:
The very large diameter of
Planet Imager
allows it to resolve planetsized detail at great distances.
36.73.
I
DENTIFY
and
S
ET
U
P
:
Follow the steps specified in the problem.
E
XECUTE
:
(a)
From the segment
,
dy
′
the fraction of the amplitude of
0
E
that gets through is
0
0
sin(
).
dy
dy
E
dE
E
kx
t
a
a
ω
′
′
⎛
⎞
⎛
⎞
⇒
=
−
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
(b)
The path difference between each little piece is
0
sin
(
sin
)
sin( (
sin
)
).
E dy
y
kx
k D
y
dE
k D
y
t
a
θ
θ
θ
ω
′
′
′
′
⇒
=
−
⇒
=
−
−
This can be rewritten as
0
(sin(
)cos(
sin
)
sin(
sin
)cos(
)).
E dy
dE
kD
t
ky
ky
kD
t
a
ω
θ
θ
ω
′
′
′
=
−
+
−
(c)
So the total amplitude is given by the integral over the slit of the above.
2
2
0
2
2
(sin(
) cos(
sin
)
sin(
sin
)
a
a
a
a
E
E
dE
dy
kD
t
ky
ky
a
ω
θ
θ
−
−
′
′
′
⇒
=
=
−
+
∫
∫
cos(
)).
kD
t
ω
−
But the second term integrates to zero, so we have:
[
]
[
]
2
2
0
0
2
2
0
0
0
sin(
sin
)
sin(
)
(cos(
sin
))
sin (
)
sin
2
sin(
(sin
) 2)
sin(
(sin
)
)
sin(
)
sin(
)
.
(sin
) 2
(sin
)
sin ...
At
0,
1
sin(
...
a
a
a
a
E
ky
E
kD
t
dy
ky
E
kD
t
a
ka
ka
a
E
E
kD
t
E
kD
t
ka
a
E
E
θ
ω
θ
ω
θ
θ
π
θ
λ
ω
ω
θ
π
θ
λ
θ
−
−
⎡
⎤
′
⎛
⎞
′
′
=
−
=
−
⎢
⎥
⎜
⎟
⎝
⎠
⎣
⎦
⎛
⎞
⎛
⎞
⇒
=
−
=
−
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
=
=
⇒
=
∫
).
kD
t
ω
−
(d)
Since
2
2
2
0
0
sin(
(sin
)/2)
sin(
2)
,
(sin
)/2
2
ka
I
E
I
I
I
ka
θ
β
θ
β
⎛
⎞
⎛
⎞
∝
⇒
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
where we have used
2
2
0
0
sin (
).
I
E
kx
t
ω
=
−
E
VALUATE
:
The same result for
( )
I
θ
is obtained as was obtained using phasors.
36.74.
I
DENTIFY
and
S
ET
U
P
:
Follow the steps specified in the problem.
E
XECUTE
:
(a)
Each source can be thought of as a traveling wave evaluated at
x
R
=
with a maximum amplitude
of
0
.
E
However, each successive source will pick up an extra phase from its respective pathlength to point
sin
2
d
P .
θ
φ
π
λ
⎛
⎞
=
⎜
⎟
⎝
⎠
which is just
2 ,
π
the maximum phase, scaled by whatever fraction the path difference,
sin
,
d
θ
is of the wavelength,
λ
. By adding up the contributions from each source (including the accumulating
phase difference) this gives the expression provided.
(b)
(
)
cos(
)
sin(
).
i kR
t
n
e
kR
t
n
i
kR
t
n
ω
φ
ω
φ
ω
φ
−
+
=
−
+
+
−
+
The real part is just cos
(
).
kR
t
n
ω
φ
−
+
So,
1
1
(
)
0
0
0
0
Re
e
cos(
).
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 Spring '06
 Buchler
 Physics, Special Relativity, Diffraction, Sin, ΔT

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