46
2 Theory
All nondiffracting beams must satisfy the condition that their transverse spatial frequency components are the same [ 16 ], that is, the general Fourier spectrum of a nondiffracting beam is located on a circular ring in the corresponding Fourier plane. In our scheme, the annular slit modulated by the radial grating with complex amplitude information is placed at the focal plane in front of the lens. Hence, the Fourier spectrum of a nondiffracting beam can conveniently be written in polar coordinates as:
U 0 ðv; uÞ ¼ AðuÞdðv � v 0 Þ; ð1Þ
where m is a radial spatial frequency, and d is the Dirac delta function that can limit the spectral distribution to a circular ring with radius m 0. A( u) gives the complex Fourier spectrum on this circular ring as a function of the azimuthal angle u. In the extended Durnin’ s experimental setup, the Fourier spectrum U 0( v, u) can be regarded as the transmission function of modulating element. In our work, A( u) is a complex amplitude function of the radial grating that resides on an annular slit of radius v 0, and its Fourier expansion function can be written as:
AðuÞ ¼ X1 m¼�1 c m exp ðimuÞ; ð2Þ
where c m is mth Fourier series coefficient, m is the number of spokes of the grating. On the basis of the diffraction integral in cylindrical coordinates, the light field distribution in the front surface of the Fourier transforming lens can be written as:
U 1 ðq 0; h 0 Þ ¼ 1 ZZ 1
U 0 ðv; uÞexp ik
� v 2 þ q 0 2 ikf �1
2f
exp � ik ð3Þ f q0 v cosðu � h 0 Þ vdvdu;
where k = 2p / k is the wave number, and k is the wavelength of the light beam. U( q 0, h 0) represents the complex amplitude distribution of the light field in the front surface of Fourier lens, q 0 and h 0 are the radial distance and azimuth angle of the beam in the front surface of Fourier lens, respectively, and f is the focal length.
In accordance with the following Jacobi-Anger expression [ 17 ]: exp ðiz cos hÞ ¼ X1 n¼�1
J. Eur. Opt. Society-Rapid Publ. 21, 35( 2025)
i n J n ðÞexpðinhÞ z; ð4Þ
U 1 ðq 0; h 0 Þ ¼ 1 ikf exp ik
2
2f q0
Z 1
0
Z 2p
0
dðv � v 0 Þexp ik
2f v2
X 1
m¼�1 The integrals of
J n
c m exp ðimuÞ X1
Z 2p
0 n¼�1
k f q0 v vdv; ð6Þ
ð�iÞ n exp ½ �inðu � h 0 ÞŠdu:
exp ½ im� ð nÞuŠdu ¼ 2pd m; n and Dirac delta function d is used, when m = n, we can get:
Z 2p
0
X 1
m¼�1
Z 1
0 c m exp ðimuÞ X1
¼ 2p X1 m¼�1 n¼�1
dðv � v 0 Þexp ik
¼ exp
ik
2f v2 0 ð�iÞ n exp ½ �inðu � h 0 ÞŠdu
ð�1Þ m c m exp ðimh 0 Þ ð7Þ
J m
2f v2
J n k f q0 v 0
k f q0 v
v 0: vdv Thus the light field distribution in the front surface of the Fourier transform lens can be obtained:
U 1 ðq 0; h 0 Þ ¼ 2p ikf exp ik
2f q02 þ v 2
0
X1 m¼�1
ð�1Þ m c m exp ðinh 0 k
ÞJ m f q0 v 0
v 0:
Ignoring the constants that do not affect the light intensity distribution, the field distribution through the lens is:
U 1 ðq 0; h 0 Þ ¼ U 1 ðq 0; h 0
2 Þexp �ik q0: ð10Þ 2f ð8Þ
ð9Þ
In the plane of( q, h, z) after the lens, the Fresnel diffraction integral of the light field distribution is:
Uðq; h; zÞ ¼ 1 ZZ 1
ikf expðikzÞ U 0 1ð q0; h 0 Þexp ik
�
�1
2z q2 þ q 0 2
exp � ik f q0 q cosðh 0 � hÞ q 0 dq 0 dh: ð11Þ where J n is the nth first type of Bessel function. One can obtain:
exp � ik qvcos u � h f ð Þ ¼
X 1 ð�iÞ n k
J n f qv exp ½ �inðu � hÞŠ: ð5Þ n¼�1
Equations( 1),( 2) and( 5) are substituted into equation( 3), one can obtain:
Substituting U 0 1ð q0; h 0 Þ into equation( 11) yields, there is:
Uðq; h; zÞ ¼ 2p k 2 zf
X1 m¼�1
Z 1 ð�iÞ m c m v 0
expðikzÞ exp ik
2f v2 0 exp ik
2f q2
0
J m
k v f q0 0 exp
ik
2 2z q0 q 0 dq 0
Z 2p
exp ðimh 0 Þexp � ik
0 f q0 q cosðh 0 � hÞ dh 0: ð12Þ