Practice problem: Find the Fraunhofer diffraction pattern of a triangular
aperture as shown in the following figure. The edges of the triangle are
expressed at x=a, y=x, and y=-x, respectively. The screen is placed at Z=Zo.
Solution: In this case, the aperture along the y direction depends on the position
x. so we may integrate first along the y direction, and then along x-direction
in the next. x
exp (ikz) a
£(kx ' ky ) =
exp (-ikx x)
exp -iky y dy dx
z
0
_x
exp iky x - exp -iky x
exp (ikz) a
exp (-ikx x)
dx
£(kx ' ky ) =
z
iky
0
£(kx ' ky ) =
a
exp (-i(kx - ky )x) dx =
f
a
f
exp (-i(kx + ky )x) dx =
£(kx ' ky ) =
exp (ikz)
iky z
a
0
exp (-i(kx - ky )x) - exp (-i(kx + ky )x) dx
1 - exp (-i(kx - ky )a)
i(kx - ky )
1 - exp (-i(kx + ky )a)
i(kx + ky )
= 2aexp (-i
= 2aexp (-i
(kx - ky )a
2
(kx + ky )a
2
)sinc(
)sinc(
(kx - ky )a
(kx - ky )a
2a exp ikz
smmc
[exp -i
2
iky z
2n
(kx + ky )a
(kx + ky )a
- exp -i
smmc
]
2
2n
1
(kx - ky )a
2n
(kx + ky )a
2n
)
)
Alternative solution:
exp (ikz)
£(kx ' ky ) =
z
1) you may verify that:
x
_x
a
0
exp (-ikx x)
a
exp (ikz)
z
0
2) again, we can see that:
a
0
O
exp -iky y dy =
£(kx ' ky ) =
exp (-ikx x)dx =
2exp (ikz)
£(kx ' ky ) =
z
x
_O
_x
rect(
2 x smmc
y
) exp -iky y dy
2x
xky
exp (-ikx x)dx
n
a
x-2
O
�ec�(
_O
exp -iky y dy dx
a
)exp (-ikx x)dX
a
x-2
Xky
sgn(x)xsinc
rect(
)exp (-ikx x)dX
n
a
-0
O
a
Using shift theorem for
£(kx ' ky ) =
O
_O
=
2 aexp ikz
z
xsinc
_i a
ky akx
£ (kx ' ky ) =
(_
O
rect( a Z)exp
_O
a
akx
exp (-ikx )sinc(
) ®
2
2n
xky
�
exp (-ikx x)dx = -i
n
�kx
rect
kx
ky
=
_i
ky
a
akx
2
2n
(-ikx x)dx= aexp (-ikx )sinc(
O
_O
O
-0
sgn(x)xsinc
s��c
xky
exp (-ikx x)dx
2n
xky
exp (-ikx x)dx
n
8 kx - ky - 8 kx + ky
a
ak
- i2aexp i��
exp (-ikx )s��c( x ) ® 8 kx - ky - 8 kx + ky
2
2n
zky
£ (kx ' ky ) =
a kx - ky
- i2aexp i��
a
exp -i kx - ky
s��c
zky
2
2n
a kx + ky
a
- exp -i kx + ky
s��c
2
2n
2
)
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2.71 / 2.710 Optics
Spring 2014
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