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Applied Optics

Applied Optics

APPLICATIONS-CENTERED RESEARCH IN OPTICS

  • Vol. 17, Iss. 6 — Mar. 15, 1978
  • pp: 964–968

Included power for obscured circular pupils

Virendra N. Mahajan  »View Author Affiliations


Applied Optics, Vol. 17, Issue 6, pp. 964-968 (1978)
http://dx.doi.org/10.1364/AO.17.000964


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Abstract

Power contained in a square area of an image formed by a diffraction-limited imaging system with a centrally obscured circular pupil is calculated and compared with the power contained in a circular area. It is shown that, regardless of the amount of obscuration, the difference between the corresponding ensquared and encircled powers is less than 9% of the total image power. Approximate expressions are obtained for the power lying outside a large square or a circular area.

© 1978 Optical Society of America

History
Original Manuscript: July 29, 1977
Published: March 15, 1978

Citation
Virendra N. Mahajan, "Included power for obscured circular pupils," Appl. Opt. 17, 964-968 (1978)
http://www.opticsinfobase.org/ao/abstract.cfm?URI=ao-17-6-964


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References

  1. Lord Rayleigh, Philos. Mag. 11, 5 (1881);Scientific Papers (Dover, New York, 1964), Vol. 1, p. 513.
  2. W. T. Welford, J. Opt. Soc. Am. 50, 749 (1960). [CrossRef]
  3. A. T. Young, Appl. Opt. 9, 1874 (1970). [PubMed]
  4. B. L. Mehta, Appl. Opt. 13, 736 (1974). [CrossRef] [PubMed]
  5. H. F. A. Tschunko, Appl. Opt. 13, 1820 (1974). [CrossRef] [PubMed]
  6. N. M. Weiner, Opt. Eng. 13, 87 (1974). [CrossRef]
  7. M. Born, E. Wolf, Principles of Optics (Pergamon, New York, 1975), p. 416.
  8. In dimensional units (as opposed to dimensionless units used in the text), if I0 is the irradiance at the pupil, the total power in the image is Pt = πD2 (1 − ∊2)I0/4, and the central irradiance is I(0;∊) = π(1 − ∊2)Pt/4λ2F2 = [πD(1 − ∊2)/4λF]2I0.
  9. This was suggested by the late Ted Edelbaum.
  10. G. N. Watson, A Treatise on the Theory of Bessel Functions (Cambridge U. P., New York, 1966), p. 195.

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