Abstract

When using the general Sweatt model for ray tracing diffractive optical elements (binary optics, holographic optical elements, etc.), lens designers must choose an index of refraction that is sufficiently high to avoid significant errors in the ray trace. We derived two new criteria for a sufficiently high index of refraction. To avoid significant errors, the index must satisfy these two new criteria in addition to Sweatt’s originally suggested criterion. In almost all cases the new criteria require a higher index than Sweatt’s original criterion.

© 1992 Optical Society of America

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References

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  1. W. C. Sweatt, “Mathematical equivalence between a holographic optical element and an ultra-high index lens,” J. Opt. Soc. Am. 69, 486–487 (1979).
    [CrossRef]
  2. For example, see W. T. Welford, “Vector raytracing equation for hologram lenses of arbitrary shape,” Opt. Commun. 14, 322–323 (1975).
    [CrossRef]
  3. W. C. Sweatt, “Describing holographic optical elements as lenses,” J. Opt. Soc. Am. 67, 803–808 (1977).
    [CrossRef]
  4. M. Born, E. Wolf, Principles of Optics (Pergamon, New York, 1980), p. 469.

1979

1977

1975

For example, see W. T. Welford, “Vector raytracing equation for hologram lenses of arbitrary shape,” Opt. Commun. 14, 322–323 (1975).
[CrossRef]

Born, M.

M. Born, E. Wolf, Principles of Optics (Pergamon, New York, 1980), p. 469.

Sweatt, W. C.

Welford, W. T.

For example, see W. T. Welford, “Vector raytracing equation for hologram lenses of arbitrary shape,” Opt. Commun. 14, 322–323 (1975).
[CrossRef]

Wolf, E.

M. Born, E. Wolf, Principles of Optics (Pergamon, New York, 1980), p. 469.

J. Opt. Soc. Am.

Opt. Commun.

For example, see W. T. Welford, “Vector raytracing equation for hologram lenses of arbitrary shape,” Opt. Commun. 14, 322–323 (1975).
[CrossRef]

Other

M. Born, E. Wolf, Principles of Optics (Pergamon, New York, 1980), p. 469.

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Figures (3)

Fig. 1
Fig. 1

Ray tracing the general Sweatt model.

Fig. 2
Fig. 2

Phased Fresnel lens.

Fig. 3
Fig. 3

Effect of variables a, λ/λ0; b, R0; c, α; d, α′.

Equations (35)

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sin θ = sin θ - λ 1 2 π d ϕ d x ,
x = x .
d t d x = λ 0 n 0 - 1 1 2 π d ϕ d x ,
t ( x ) = λ 0 n 0 - 1 ϕ ( x ) 2 π + t 0 ,
n ( λ ) = ( n 0 - 1 ) λ + λ 0 λ 0 .
n sin ( + δ ) = sin ( θ + δ ) ,
n sin ( - δ ) = sin ( θ - δ ) .
sin θ = sin θ - n ( δ + δ ) + δ cos θ + δ cos θ .
δ + δ - δ = d t d x .
sin θ = sin θ - λ ( 1 2 π ) d ϕ d x - d t d x ( 1 - cos θ ) .
x = x + t tan x + t n sin θ .
Δ ( sin θ ) = | d t d x ( 1 - cos θ ) | .
Δ θ Δ ( sin θ ) / cos θ = | d t d x ( sec θ - 1 ) | .
Δ x = | t n sin θ | .
ϕ ( x ) = 2 π λ 0 [ f - ( x 2 + f 2 ) 1 / 2 ] ,
α = arctan ( R / f ) .
t 0 = f n 0 - 1 ( sec α - 1 ) .
max | d t d x | = 1 n 0 - 1 sin α ,
max sec θ - 1 = sec ( α + α ) - 1.
max Δ θ 1 n 0 - 1 [ sec ( α + α ) - 1 ] sin α .
max Δ θ 0.50 λ / R .
n 0 - 1 1 0.50 ( λ 0 λ ) R λ 0 [ sec ( α + α ) - 1 ] sin α .
max t ( x ) = t 0 = f n 0 - 1 ( sec α - 1 ) ,
max sin θ = sin α ,
max Δ x f λ 0 ( n 0 - 1 ) [ ( n 0 - 1 ) λ + λ 0 ] × sin α ( sec α - 1 ) .
max Δ x 0.50 λ f / R .
n 0 - 1 [ 1 0.50 ( λ 0 λ ) 2 R λ 0 sin α ( sec α - 1 ) ] 1 / 2 .
Δ ( OPD ) = n t ( sec - 1 ) n t 2 / 2 t sin 2 θ / 2 n .
max Δ ( OPD ) f λ 0 2 ( n 0 - 1 ) [ ( n 0 - 1 ) λ + λ 0 ] × sin 2 α ( sec α - 1 ) .
max Δ ( OPD ) λ / 50.0
n 0 - 1 [ 50.0 2 ( λ 0 λ ) 2 R λ 0 sin 2 α ( csc α - cot α ) ] 1 / 2 .
λ 0 / λ = 1.0 ,
R / λ 0 = 20000 ,
α = 5 ° ,
α = 9.5 ° .

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