Abstract

Exact formulas are derived for the field of view of an optical instrument with rectangular primary and secondary apertures and a rectangular field stop. The formulas derived include the effects of diffraction at the primary and secondary apertures. Numerical results are presented for an instrument with square optics, viewing a uniform monochromatic source. Similar calculations have been done for optical instruments with circular apertures by Goldberg and McCulloch [ Appl. Opt. 8, 1451 ( 1969)]. The calculations for an instrument with square optics without a secondary aperture show that if the angle θf subtended by the geometrically determined field of view is 35 times as large as the ratio of wavelength to aperture size, then less than 5% of the energy collected at the detector will be from areas of the source outside the geometrical field of view. Using these formulas, an expression has been derived for the field of view of the same type of instrument viewing an infinite monochromatic source having two uniform brightness levels separated by a straight edge. Numerical results are presented for an instrument with square optics. If θf is at least ten times as large as the ratio of wavelength to aperture size, than 90% of the instrument’s response to the step change in brightness occurs while the step is crossing the geometrically determined field of view.

© 1970 Optical Society of America

Full Article  |  PDF Article

References

  • View by:
  • |
  • |
  • |

  1. I. R. Goldberg, A. W. McCulloch, Appl. Opt. 8, 1451 (1969).
    [CrossRef] [PubMed]
  2. M. Born, E. Wolf, Principles of Optics (Pergamon Press, New York, 1959), Chap. 8.
  3. F. B. Hildebrand, Advanced Calculus for Engineers (Prentice-Hall, New York, 1948), Chaps. 8 and 9.

1969 (1)

Born, M.

M. Born, E. Wolf, Principles of Optics (Pergamon Press, New York, 1959), Chap. 8.

Goldberg, I. R.

Hildebrand, F. B.

F. B. Hildebrand, Advanced Calculus for Engineers (Prentice-Hall, New York, 1948), Chaps. 8 and 9.

McCulloch, A. W.

Wolf, E.

M. Born, E. Wolf, Principles of Optics (Pergamon Press, New York, 1959), Chap. 8.

Appl. Opt. (1)

Other (2)

M. Born, E. Wolf, Principles of Optics (Pergamon Press, New York, 1959), Chap. 8.

F. B. Hildebrand, Advanced Calculus for Engineers (Prentice-Hall, New York, 1948), Chaps. 8 and 9.

Cited By

OSA participates in CrossRef's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.

Alert me when this article is cited.


Figures (6)

Fig. 1
Fig. 1

Geometry of the coordinate systems for Eq. (1).

Fig. 2
Fig. 2

Fraction of energy received at the detector [defined by Eq. (9)] as a function of Γ,W,Z [defined by Eq. (6)] and α, the ratio of the sizes of the secondary to primary apertures, for α = 0.0, 0.5, and W = Z = 1.0.

Fig. 3
Fig. 3

Fraction of energy received at the detector as a function of Γ, W, Z, and α for α = 0.0, 0.5, and W = Z = 1.5.

Fig. 4
Fig. 4

Fraction of energy received at the detector as a function of Γ, W, Z, and α, for α = 0.0, 0.5, and W = Z = 2.0.

Fig. 5
Fig. 5

Fraction of energy received at the detector as a function of Γ, W, Z, and α, for α = 0.0, 0.5, and W = Z = 4.0.

Fig. 6
Fig. 6

Response to a single step source of an optical instrument with a square primary aperture and no secondary aperture [Eq. (15)]. Instrument scan is perpendicular to the edge of the step, which is located at Z1 = 0.

Tables (2)

Tables Icon

Table I Values of W, Γ, and α for Which the Ratios Defined by Eqs. (6) and (9) Were Computed

Tables Icon

Table II Values of F(W,W,α) Defined by Eq. (9) for the Instruments Described in the Text for the Angular Limits Listed Below

Equations (21)

Equations on this page are rendered with MathJax. Learn more.

I ( x x , y y ) d x d y = A s Ω s d x d y λ 2 f 4 × sinc 2 ( 2 π x x λ f X 0 ) sinc 2 ( 2 π y y λ f Y 0 ) ,
d E ( x , y ) recevied = d x d y X 1 X 1 d x Y 1 Y 1 d y I ( x x , y y ) .
E recevied ( X 1 , Y 1 ) = X 1 X 1 d x Y 1 Y 1 d y X 1 X 1 d x × Y 1 Y 1 d y I ( x x , y y ) .
E recevied ( X L , Y L ) = X L X L d x Y L Y L d y X 1 X 1 d x × Y 1 Y 1 d y I ( x x , y y ) .
F ( X 1 , Y 1 ) = 1 E received ( X 1 , Y 1 ) E received ( , ) .
F ( W , Z = W ) = 1 ( Γ / 2 π ) Q ( W , Γ ) ,
W = X L / X 1 , Z = Y L / Y 1 = X L / X 1 = W , Γ = 2 π X 0 X 1 / λ f .
Q ( W , Γ ) = 4 W 0 1 d w sinc 2 Γ w + 2 0 W d w ( W w ) [ sinc 2 Γ ( w + 1 ) sinc 2 Γ ( w 1 ) ] ,
w = ( x x ) / x 1 .
I ( x x , y y ) d x d y = A c s Ω s d x d y λ 2 f 4 X { sinc [ 2 π ( x x ) X 0 λ f ] sinc [ 2 π ( y y ) Y 0 λ f ] α 2 sinc [ 2 π ( x x ) λ f X 0 ] sinc [ 2 π ( y y ) Y 0 λ f ] } 2 .
F ( W , Z = W , α ) = 1 Q 2 ( W , Γ ) + α 4 Q 2 ( W , α Γ ) 2 α 2 T 2 ( W , Γ ) ( 4 π 2 / Γ 2 ) ( 1 α 2 ) .
T ( W , Γ ) = 4 W 0 1 d w sinc ( Γ w ) sinc ( α Γ w ) + 2 0 W d w ( W w ) [ sinc Γ ( w + 1 ) sinc α Γ ( w + 1 ) sinc Γ ( w 1 ) sinc α Γ ( w 1 ) ] .
θ f = 2 X 1 / f = 2 Y 1 / f .
θ L = 2 X L / f = 2 Y L / f = W θ f ,
s / Ω s = E 0 to the left of the step , s / Ω s = E 1 to the right of the step ,
Energy received = E 0 A D f 2 ( E 0 E 1 ) A D B ( Z 1 ) 2 f 2 ,
A D = area of field stop , B ( Z 1 ) = 1 Γ π 0 1 + Z 1 d z ( 1 + Z 1 z ) sinc 2 Γ z + Γ π 0 1 Z 1 d z ( 1 Z 1 z ) sinc 2 Γ z , z = ( x x / X 1 ) ,
Z 1 = X b / X 1 .
Z 1 ( t ) = Z 1 ( 0 ) + w f t / X 1 ,
B [ Z 1 ( t ) ] = energy received E 0 A D / f 2 ( E 1 E 0 ) A D / 2 f 2 ,
Γ = 1.0,2.0,10.0,20.0,100.0 50.0 Z 1 , 5.0 , Δ Z 1 = 0.5.

Metrics