Abstract

Astronomical aspherical mirrors may be tested in the optical shop with a Hartmann null test. A null test is obtained by placing small wedges over each hole of the Hartmann screen. The wedges have an angle between the two faces such that the spherical aberration with the object and the image at the center of curvature is just compensated.

© 1972 Optical Society of America

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References

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  1. A. B. Meinel, in Applied Optics and Optical Engineering, R. Kingslake, Ed. (Academic, New York, 1970), Vol. 5, Chap. 6.
  2. D. H. Schulte, Appl. Opt. 7, 119 (1968).
    [CrossRef] [PubMed]
  3. M. V. R. K. Murty, Appl. Opt. 1, 364 (1962).
    [CrossRef]
  4. D. Malacara, “Testing of Optical Surfaces,” Ph. D. Thesis, The University of Rochester (1965).

1968 (1)

1962 (1)

Malacara, D.

D. Malacara, “Testing of Optical Surfaces,” Ph. D. Thesis, The University of Rochester (1965).

Meinel, A. B.

A. B. Meinel, in Applied Optics and Optical Engineering, R. Kingslake, Ed. (Academic, New York, 1970), Vol. 5, Chap. 6.

Murty, M. V. R. K.

Schulte, D. H.

Appl. Opt. (2)

Other (2)

D. Malacara, “Testing of Optical Surfaces,” Ph. D. Thesis, The University of Rochester (1965).

A. B. Meinel, in Applied Optics and Optical Engineering, R. Kingslake, Ed. (Academic, New York, 1970), Vol. 5, Chap. 6.

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Figures (7)

Fig. 1
Fig. 1

Hartmann test in using glass wedges.

Fig. 2
Fig. 2

Hartmann screen used on the test.

Fig. 3
Fig. 3

Orientation of the glass wedges.

Fig. 4
Fig. 4

Computation and testing of the glass wedges.

Fig. 5
Fig. 5

Tolerance in number of Fizeau fringes over glass wedge vs its diameter.

Fig. 6
Fig. 6

Glass wedge orientation error.

Fig. 7
Fig. 7

Tolerance in glass wedge orientation error.

Equations (20)

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f ( s ) = C S 2 1 + [ 1 - ( K + 1 ) C 2 S 2 ] 1 2 + A 4 S 4 + A 6 S 6 + A 8 S 8 + A 10 S 10 ,
f r ( s ) = C 0 S 2 / [ 1 + ( 1 - C 0 2 S 2 ) 1 2 ] .
g ( s ) = f ( s ) - f r ( s ) .
g ( s ) = B 2 S 2 + B 4 S 4 + B 6 S 6 + B 8 S 8 + B 10 S 10 ,
φ = d g ( s ) / d s = 2 B 2 S + 4 B 4 S 3 + 6 B 6 S 5 + 8 B 8 S 7 + 10 B 10 S 9 .
B 2 = ( C - C 0 ) / 2 ,
B 4 = ( 1 / 8 ) [ ( K + 1 ) C 3 - C 0 3 ] + A 4 ,
B 6 = ( 1 / 16 ) [ ( K + 1 ) 2 C 5 - C 0 5 ] + A 6 ,
B 8 = ( 5 / 128 ) [ ( K + 1 ) 3 C 7 - C 0 7 ] + A 8 ,
B 10 = ( 7 / 256 ) [ ( K + 1 ) 4 C 9 - C 0 9 ] + A 10 .
( φ + θ ) / θ = N ,
M = γ d / λ = N / ( N - 1 ) ,             φ d / λ .
δ M = N / ( N - 1 ) ,             ( d / λ ) δ φ .
δ φ = λ / D = λ / 20 d .
δ M = N / 20 ( N - 1 ) .
δ φ = ( 4.85 × 10 - 6 ) β ,
δ M = ( 4.85 × 10 - 6 ) ( N / ( N - 1 ) ( d / λ ) β .
Δ = 2 t sin ψ [ 1 - cos ψ / ( N 2 - sin 2 ψ ) 1 2 ] .
δ φ = ( Δ / R ) ,
R β t = 2 sin ψ 2.424 × 10 - 6 [ 1 - cos ψ N 2 - sin 2 ψ ] 120 [ ( N - 1 ) / N ] ψ ,

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