Abstract

A simplified way is proposed to compute the paraxial diffraction of a zone plate. The complicated diffraction field is approximated by an equivalent field, which is the diffraction field of a hole weighted by the area. Based on this equivalent expression of the diffraction field, a graphic method is applied to calculate the resolving power of a zone plate. The result is consistent with those reported in earlier publications [Am. J. Phys. 19, 359 (1951); J. Opt. Soc. Am. 57, 610 (1967)].

© 1994 Optical Society of America

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References

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  1. O. E. Myers, Am. J. Phys. 19, 359 (1951).
    [CrossRef]
  2. D. J. Stigliani, R. Mittra, R. G. Semonin, J. Opt. Soc. Am. 57, 610 (1967).
    [CrossRef]
  3. J. Kirz, J. Opt. Soc. Am. 64, 301 (1974).
    [CrossRef]
  4. M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980), Chap. 8.
  5. C. Harper, Introduction to Mathematical Physics (Prentice-Hall, Englewood Cliffs, N.J., 1976), Chap. 6.

1974 (1)

1967 (1)

1951 (1)

O. E. Myers, Am. J. Phys. 19, 359 (1951).
[CrossRef]

Born, M.

M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980), Chap. 8.

Harper, C.

C. Harper, Introduction to Mathematical Physics (Prentice-Hall, Englewood Cliffs, N.J., 1976), Chap. 6.

Kirz, J.

Mittra, R.

Myers, O. E.

O. E. Myers, Am. J. Phys. 19, 359 (1951).
[CrossRef]

Semonin, R. G.

Stigliani, D. J.

Wolf, E.

M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980), Chap. 8.

Am. J. Phys. (1)

O. E. Myers, Am. J. Phys. 19, 359 (1951).
[CrossRef]

J. Opt. Soc. Am. (2)

Other (2)

M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980), Chap. 8.

C. Harper, Introduction to Mathematical Physics (Prentice-Hall, Englewood Cliffs, N.J., 1976), Chap. 6.

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Figures (2)

Fig. 1
Fig. 1

Abscissa, x = 2πβr1/λ; ordinate, Uc. Positive zone plate (N = 3,199), β3 > 1.22λ/D3. The first sidelobe is smaller than that of a lens with the same aperture. β199 = 1.21λ/D199.

Fig. 2
Fig. 2

Abscissa, x = 2πβr1/λ; ordinate, Uc. Negative zone plate (N = 4,200), β4 < 1.22λ/D4. The first sidelobe is bigger than that of a lens with the same aperture. β200 = 1.22λ/D200.

Equations (7)

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U c = U 0 λ β exp ( j π F β 2 λ ) × n = 0 1 / 2 ( N 1 ) [ r 2 n + 1 J 1 ( 2 π β r 2 n + 1 λ ) r 2 n J 1 ( 2 π β r 2 n λ ) ] ,
U c = n = 1 1 / 2 ( N 1 ) [ π r 2 n + 1 2 J 1 ( 2 π β r 2 n + 1 / λ ) 2 π β r 2 n + 1 / λ + 2 π r 1 2 J 1 ( 2 π β r 1 / λ ) 2 π β r 1 / λ + π r 2 n + 1 2 J 1 ( 2 π β r 2 n + 1 / λ ) 2 π β r 2 n + 1 / λ 2 π r 2 n 2 J 1 ( 2 π β r 2 n / λ ) 2 π β r 2 n / λ + π r 2 n 1 2 J 1 ( 2 π β r 2 n 1 / λ ) 2 π β r 2 n 1 / λ π r 2 n 1 2 J 1 ( 2 π β r 2 n 1 / λ ) 2 π β r 2 n 1 / λ ] .
J 1 ( x ) x = 1 x k = 0 ( 1 ) k ( x / 2 ) 1 + 2 k k ! ( 1 + k ) ! = 1 2 1 2 ! 1 2 3 x 2 + 1 2 ! 3 ! 1 2 5 x 4 ....
ε 2 n ( β ) = π r 2 n + 1 2 J 1 ( 2 π β r 2 n + 1 / λ ) 2 π β r 2 n + 1 / λ 2 π r 2 n 2 J 1 ( 2 π β r 2 n / λ ) 2 π β r 2 n / λ + π r 2 n 1 2 J 1 ( 2 π β r 2 n 1 / λ ) 2 π β r 2 n 1 / λ π ( 2 n + 1 ) F λ [ 1 2 1 2 1 8 ( 2 π / λ ) 2 β 2 ( 2 n + 1 ) F λ + 1 12 1 32 ( 2 π / λ ) 4 β 4 ( 2 n + 1 ) 2 F 2 λ 2 ] 2 π ( 2 n ) F λ [ 1 2 1 2 1 8 ( 2 π / λ ) 2 β 2 ( 2 n ) F λ + 1 12 1 32 ( 2 π / λ ) 4 β 4 ( 2 n ) 2 F 2 λ 2 ] + π ( 2 n 1 ) F λ [ 1 2 1 2 1 8 ( 2 π / λ ) 2 β 2 ( 2 n 1 ) F λ + 1 12 1 32 ( 2 π / λ ) 4 β 4 ( 2 n 1 ) 2 F 2 λ 2 ] π F 2 λ 2 β 2 [ 1 8 ( 2 π / λ ) 2 + 1 64 ( 2 π / λ ) 4 β 2 ( 2 n ) F λ ] .
U c = π r N 2 J 1 ( 2 π β r N / λ ) 2 π β r N / λ + π r 1 2 J 1 ( 2 π β r 1 / λ ) 2 π β r 1 / λ .
U c = π r N 2 J 1 ( 2 π β r N / λ ) 2 π β r N / λ π r 1 2 J 1 ( 2 π β r 1 / λ ) 2 π β r 1 / λ ,
β 3 = 2 . 60 2 π λ r 1 = 2 . 60 3 π λ 2 3 r 1 = 1 . 43 λ D 3 ,

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