## Abstract

The general theory of optically compensated varifocal systems is applied to a three-component system composed of two movable components placed on either side of a fixed component. The two movable components are displaced in unison to change the over-all focal length of the system. The three-component system is analyzed in detail and an iteration method for the solution of the varifocal equations is developed which enables us to determine the Gaussian parameters of the system for any predetermined focal range in a matter of a few minutes (without the use of computers). Using this method, explicit expressions are found for the approximate values of the parameters of the optimum three-component systems as functions of the focal range. A numerical example is given to illustrate the iteration method.

© 1962 Optical Society of America

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### References

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1. L. Bergstein and L. Motz, J. Opt. Soc. Am. 52, 353 (1962).
[Crossref]
2. L. Bergstein, J. Opt. Soc. Am. 48, 154 (1958).
[Crossref]
3. The operating range of focal lengths of the system is determined by the “relative focal range” ℛ defined as the ratio of maximum to minimum over-all focal lengths, i.e.,R≜FmaxFmin=fmaxfmin.It was already pointed out1,2 that for any required focal range ℛ two lens types are possible. Either a lens system can be chosen which has the maximum focal length when the movable components are in the front position, or a system can be chosen which has its maximum focal length when the movable components are in the rear position. We refer to the first system as the P system; the latter is referred to as the N system. We define the “focal ratio”r≜F(0)F(1)=f(0)f(1).For the P system r=ℛ, whereas r=1/ℛ for the N system. We also introduce the normalized rangeτ≜f(0)-f(1)f(0)+f(1)=r-1r+1=±R-1R+1,where the positive sign applies to the P system, the negative sign to the N system. We note that 1.0<ℛ<∞, 0<r<∞, and −1.0<τ<+1.0.
4. For τ=0, Eq. (16) reduces to b2y(z)=z(z−z2) (z−1.0).
5. The choice between the two optimum systems is governed by a number of considerations.2 (The PNP system is generally preferable.)

#### Other (3)

The operating range of focal lengths of the system is determined by the “relative focal range” ℛ defined as the ratio of maximum to minimum over-all focal lengths, i.e.,R≜FmaxFmin=fmaxfmin.It was already pointed out1,2 that for any required focal range ℛ two lens types are possible. Either a lens system can be chosen which has the maximum focal length when the movable components are in the front position, or a system can be chosen which has its maximum focal length when the movable components are in the rear position. We refer to the first system as the P system; the latter is referred to as the N system. We define the “focal ratio”r≜F(0)F(1)=f(0)f(1).For the P system r=ℛ, whereas r=1/ℛ for the N system. We also introduce the normalized rangeτ≜f(0)-f(1)f(0)+f(1)=r-1r+1=±R-1R+1,where the positive sign applies to the P system, the negative sign to the N system. We note that 1.0<ℛ<∞, 0<r<∞, and −1.0<τ<+1.0.

For τ=0, Eq. (16) reduces to b2y(z)=z(z−z2) (z−1.0).

The choice between the two optimum systems is governed by a number of considerations.2 (The PNP system is generally preferable.)

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### Figures (12)

Fig. 1

The three-component optically compensated varifocal system.

Fig. 2

The three-component optically compensated varifocal system and its Gaussian parameters, its over-all focal length f(z) and image-plane deviation y(z) as functions of the normalized displacement z.

Fig. 3

The optimum distribution of the points (z1,z2,z3) of full compensation of a three-component varifocal system with full compensation at both ends of the operating range as a function of the focal range τ. z1=0, z2=0.5−, z3=1.0.

Fig. 4

The value of β2y3(max) as a function of the focal range τ; y3(max) is the maximum value of the normalized image-plane deviation of a three-component varifocal system with full compensation of both ends of the operating range and an optimized image deviation function, β2≡(r−1)b2=[2τ/(1−τ)]b2, and b2 is a Gaussian bracket of the system. The image-plane deviation y(z) as a function of the displacement z of the movable components is shown above the graph of β2y3(max).

Fig. 5

The parameters f1, f22, and β2 of three-component varifocal systems as functions of the parameter λ, or of l; f1 is the focal length of the rear movable component, f2 is the focal length of the stationary component, and β2≡(r−1)b2=[2τ/(1−τ)]b2, where b2 is a Gaussian bracket of the system and τ is the normalized focal range. The index P refers to P systems, the index N to N systems. (l)P=l1′(0) and (l)N=−l1′(1), where l1′(0) and l1′(1) are the normalized final image distances in the two extreme positions (0 and 1.0) of the movable components; λ=l1′(0)−0.5−(m)=l1′(1)+0.5−(m), where m≡1/4(1/τ+2) and, =0.5−z2 is given by Eq. (20) or Fig. 3. (τ, m and are positive for P systems and negative for N systems.)

Fig. 6

Possible three-component varifocal systems. The systems shown in (Pa), (Pb), and (Pc) are P systems, the systems shown in (Na), (Nb), (Nc), and (Nd) are N systems. Also shown is the focal range of these systems.

Fig. 7

The values of the parameters (l)0, λ0, and (β2)0 of the optimum three-component varifocal systems as functions of the focal range τ; (l)0, λ0 and (β2)0 are the values of l, λ and β2≡[2τ/(1−τ)]b2 for systems in which the distance s21 between the principal planes of the stationary component and the rear movable component in the position z=0 is zero.

Fig. 8

The normalized focal lengths (f1)0, (f2)0 and, (f3)01 of the components, the final image distance [l1′(0)]0 and the normalized maximum over-all focal length (fmax)01 of the optimum three-component varifocal systems as functions of the focal range τ. The index 0 indicates that the particular parameter was found under the assumption that s21=0, the index 01, that the parameter in question was found under the assumption that s21=0 and s32=1.0; s21 and s32 are the distances between the principal planes of the components 2 and 1, and 3 and 2, respectively, in the position z=0 of the movable components.

Fig. 9

The maximum value y3(max)0 of the normalized image-plane deviation of the optimum four-component varifocal systems with full compensation at both ends of the operating range (and an optimized image deviation function) as a function of the focal range τ. The distance s21 between the principal planes of the stationary and rear movable components in the position z=0 is assumed to be zero.

Fig. 10

The maximum value YT(max)01 of the image-plane deviation of optimum three-component varifocal systems with full compensation at both ends of the operating range (and an optimized image deviation function) as a function of the focal range τ. The distances between the principal planes of the components in the position z=0 are assumed to be s21=0 and s32=1.0, respectively. FT(max) is the maximum over-all focal length of the system and Zm is the maximum displacement of the movable components.

Fig. 11

The distance s21 between the principal planes of the stationary and rear movable components as a function of the parameter l for the two three-component systems of the example of Sec. III.

Fig. 12

The two optimum three-component systems of the example of Sec. III, and their image-plane deviation and focal range.

### Equations (99)

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$f k ≜ F k Z m , d k , k - 1 ≜ D k , k - 1 Z m , s k , k - 1 ≜ S k , k - 1 Z m ,$
$z = Z / Z m .$
$ψ 1 ( 3 , 1 ) ≡ b 1 = - d 32 + d 21 , ψ 2 ( 3 , 1 ) ≡ b 2 = - d 32 d 21 + f 2 2 , ψ 1 ( 3 , 2 ) ≡ a 1 = - d 32 .$
$f ( z ) = - f 1 f 2 f 3 / ( z 2 + b 1 z + b 2 ) .$
$x 1 ′ ( z ) = f 1 2 z + a 1 z 2 + b 1 z + b 2 ,$
$y ( z ) = z 3 + c 1 z 2 + c 2 z + c 3 z 2 + b 1 z + b 2 ,$
$c 1 = b 1 - ( x ′ - y 0 ) , c 2 = b 2 - ( x ′ - y 0 ) b 1 + f 1 2 , c 3 = y 0 b 2 ;$
$x ′ ≡ x 1 ′ ( 0 ) = f 1 2 a 1 / b 2 .$
$y ( z ) = ( z - z 1 ) ( z - z 2 ) ( z - z 3 ) z 2 + b 1 z + b 2 = z 3 - γ 1 z 2 + γ 2 z - γ 3 z 2 + b 1 z + b 2 ,$
$γ 1 = z 1 + z 2 + z 3 , γ 2 = z 1 z 2 + z 1 z 3 + z 2 z 3 , γ 3 = z 1 z 2 z 3 .$
$γ 1 = 1 + z 2 , γ 2 = z 2 , γ 3 = 0.$
$b 1 - ( x ′ - y 0 ) = - γ 1 , b 2 - ( x ′ - y 0 ) b 1 + f 1 2 = γ 2 ,$
$y 0 = - γ 3 / b 2 .$
$( r - 1 ) b 2 = 1 + b 1 .$
$f 2 + d 21 + f 1 ≡ s 21 > s 21 ( min ) ,$
$f 3 = s 32 - d 32 - f 2 ,$
$and s 21 ( min ) ≈ 0 , s 32 ( min ) ≈ 1.0.$
$∣ y ( 0 ) ∣ = ŷ 1 , 2 = ŷ 2 , 3 = ∣ y ( 1 ) ∣ ,$
$ŷ 1 , 2 = ŷ 2 , 3 ,$
$y ( z ) ≈ 1 b 2 ( z - z 1 ) ( z - z 2 ) ( z - z 3 ) 1 + ( r - 1 ) z = [ 1 ( r - 1 ) b 2 ] ( z - z 1 ) ( z - z 2 ) ( z - z 3 ) ( 1 - τ ) / 2 τ + z .$
$y ( z ˜ ) ≈ y ( 1 - z ) ≈ [ 1 ( r - 1 ) b 2 ] ( z ˜ - z ˜ 1 ) ( z ˜ - z ˜ 2 ) ( z ˜ - z ˜ 3 ) - ( 1 + τ ) / 2 τ + z ˜ .$
$[ ( r - 1 ) b 2 ] y P ( z ) ≈ [ ( r - 1 ) b 2 ] y N ( 1 - z ) .$
$( z i ) N ≈ 1 - ( z 4 - i ) P , i = 1 , 2 , 3.$
$z ˆ i , i + 1 ≈ 1 2 ( z i + z i + 1 ) .$
$[ ( r - 1 ) b 2 ] y ( z ) ≈ z ( z - z 2 ) ( z - 1.0 ) ( 1 + τ ) / 2 τ + z .$
$z 2 ≡ 0.5 - ∊ ,$
$z ˆ 1 , 2 ≈ 1 2 z 2 = 0.25 - 1 2 ∊ , z ˜ 2 , 3 ≈ 1 2 ( 1 + z 2 ) = 0.75 - 1 2 ∊ ,$
$[ ( r - 1 ) b 2 ] ŷ 1 , 2 ≈ ( ∣ τ ∣ 32 ) ( 1 - 2 ∊ ) 2 ( 3 + 2 ∊ ) 1 - 1 2 τ - τ ∊ ,$
$[ ( r - 1 ) b 2 ] ŷ 2 , 3 ≈ ( ∣ τ ∣ 32 ) ( 1 + 2 ∊ ) 2 ( 3 - 2 ∊ ) 1 + 1 2 τ - τ ∊ .$
$3 τ - 20 ∊ + 24 τ ∊ 2 + 16 ∊ 3 - 16 τ ∊ 4 = 0.$
$∊ ≈ 0.1508 τ ( 1 + 0.17 τ 2 + 0.15 τ 4 ) .$
$[ ( r - 1 ) b 2 ] y ( max ) = [ ( r - 1 ) b 2 ] 1 2 ( ŷ 1 , 2 + ŷ 2 , 3 ) ≈ ( 3 3 2 ∣ τ ∣ ) 1 - ( 8 / 3 ) τ ∊ + ( 4 / 3 ) ∊ 2 1 - 0.25 τ 2 - 2 τ ∊ + τ 2 ∊ 2 .$
$[ ( r - 1 ) b 2 ] y ( max ) ≈ 0.094 ∣ τ ∣ ( 1 + 0.146 τ 2 + 0.158 τ 4 ) .$
$d 32 - d 21 - d 32 f 1 2 - γ 3 f 2 2 - d 32 d 21 = γ 1 , f 2 2 - d 32 d 21 - ( d 32 - d 21 ) d 32 f 1 2 - γ 3 f 2 2 - d 32 d 21 + f 1 2 = γ 2 , ( r - 1 ) ( f 2 2 - d 32 d 21 ) + d 32 - d 21 = 1 , f 2 + d 21 + f 1 = s 21 ,$
$- b 1 + x ′ + γ 3 / b 2 = γ 1 , b 2 - x ′ b 1 - γ 3 b 1 / b 2 + f 1 2 = γ 2 , ( r - 1 ) b 2 - b 1 = 1.0.$
$b 1 = ( r - 1 ) b 2 - 1.0 , x ′ = γ 1 + b 1 - γ 3 / b 2 , f 1 2 = γ 2 + γ 1 b 1 + b 1 2 - b 2 .$
$d 32 = a 1 = - b 2 x ′ / f 1 2 , d 21 = b 1 + d 32 , f 2 2 = b 2 + d 32 d 21 .$
$[ ∣ ( 1 / 2 τ + ∊ ) ∣ ] < ( r - 1 ) b 2 < ∞ ) .$
$s 21 = f 2 + d 21 + f 1 .$
$f max = - ( 2 τ 1 - ∣ τ ∣ ) f 1 f 2 f 3 ( r - 1 ) b 2 .$
$- b 1 + x ′ = γ 1 = 1.5 - ∊ , b 2 - x ′ b 1 + f 1 2 = γ 2 = 0.5 - ∊ , ( r - 1 ) b 2 - b 1 = 1.$
$l = { l 1 ′ ( 0 ) = x ′ + f 1 , for the P system , - l 1 ′ ( 1 ) = - ( x ′ + f 1 - 1.0 ) , for the N system .$
$λ ≜ l 1 ′ ( 0 ) - 1 2 [ r / ( r - 1 ) + z 2 ] = ± [ l - 1 2 - ∣ ( m - ∊ ) ∣ ] ,$
$m ≡ 1 2 [ r / ( r - 1 ) - z 2 ] = 1 4 ( 1 / τ + 2 ∊ ) .$
$f 1 = ( λ 2 - m 2 ) / 2 λ ,$
$x ′ = l 1 ′ ( 0 ) - f 1 = l 1 ′ ( 1 ) - f 1 + 1.0 = ( λ + m ) 2 / 2 λ + 1 2 - ∊ ,$
$β 2 ≡ ( r - 1 ) b 2 = x ′ - ( 1 2 - ∊ ) = ( λ + m ) 2 / 2 λ ,$
$b 1 = β 2 - 1.0.$
$f 2 2 = 4 ( m - ∊ ) λ ( λ + m ) 2 - ( 1 - 4 ∊ 2 ) λ 2 ( λ - m ) 4 .$
$or ( λ ) P > 0 and ( - λ ) N > 0 , l > ( 3 ℛ - 1 ) / 4 ( ℛ - 1 ) - 1 2 ∣ ∊ ∣ .$
$( λ ) P ( - λ ) N } > ∣ m ∣ = 1 4 ( ℛ + 1 ℛ - 1 + 2 ∣ ∊ ∣ ) , or , l > ℛ ℛ - 1 .$
$s 21 = 1 2 + { λ + m - 1 2 τ + ( 1 - τ 2 2 τ 2 ) λ ( λ - m ) 2 × [ 1 - [ 1 + 2 τ 2 ( 1 - τ 2 ) ( 1 - 2 m τ ) ( λ - m ) 2 τ λ ] 1 2 ] } .$
$s 21 = s 21 ( λ , τ ) = 0 ,$
$λ 0 ≈ 1.453 τ [ ( 1 - 0.293 τ 2 - 0.192 τ 4 ) + 0.172 τ ( 1 - 0.14 τ 2 - 0.03 τ 4 ) ] ,$
$( l ) 0 ≈ 1.703 ∣ τ ∣ [ ( 1 - 0.294 τ 2 - 0.171 τ 4 ) + 0.147 τ ( 1 - 0.14 τ 2 - 0.03 τ 4 ) ] + 0.5 ,$
$( β 2 ) 0 ≈ 0.999 τ [ ( 1 - 0.135 τ 2 - 0.062 τ 4 ) + 0.121 τ ( 1 - 0.13 τ 2 - 0.17 τ 4 ) ] .$
$and ( l ) 0 P < l = l 1 ′ ( 0 ) < ∞ , for the P N P system , ℛ ℛ - 1 < l = - l 1 ′ ( 1 ) < ( l ) 0 N , for the N P N system ,$
$( f 3 ) 01 ≡ ( f 3 ) 0 - ( s 32 - 1.0 ) = 1.0 - ( d 32 ) 0 - ( f 2 ) 0$
$( f 1 ) 0 ≈ + 0.704 τ [ ( 1 - 0.313 τ 2 - 0.267 τ 4 ) + 0.184 τ ( 1 - 0.14 τ 2 + 0.10 τ 4 ) ] ,$
$( f 2 ) 0 ≈ - 0.704 τ [ ( 1 - 0.277 τ 2 - 0.239 τ 4 ) - 0.184 τ ( 1 - 0.01 τ 2 + 0.32 τ 4 ) ] ,$
$( f 3 ) 01 ≈ + 1.704 τ [ ( 1 - 0.210 τ 2 - 0.146 τ 4 ) + 0.147 τ ( 1 - 0.10 τ 2 - 0.06 τ 4 ) ] .$
$[ l 1 ′ ( 0 ) ] 0 ≈ 1.703 τ [ ( 1 - 0.294 τ 2 - 0.171 τ 4 ) + 0.440 τ ( 1 - 0.05 τ 2 - 0.01 τ 4 ) ] .$
$( f max ) 01 = - [ 2 τ / ( 1 - ∣ τ ∣ ) ] [ ( f 1 ) 0 ( f 2 ) 0 ( f 3 ) 01 / ( β 2 ) 0 ] .$
$( f max ) 01 ≈ 2.075 τ [ ( 1 + 0.884 τ 2 + 0.808 τ 4 ) + 0.022 τ ( 1 + 2.34 τ 2 - 3.36 τ 4 ) ] .$
$y ( max ) 0 = [ β 2 y ( max ) ] / ( β 2 ) 0 .$
$Y T ( max ) 01 F T 2 ( max ) / Z m ≡ Θ ( max ) 01 = y ( max ) 0 ( f max ) 01 2 = ( 1 - ∣ τ ∣ ) 2 4 τ 2 × ( ( β 2 ) 0 [ β 2 y ( max ) ] [ ( f 1 ) 0 ( f 2 ) 0 ( f 3 ) 01 ] 2 ) .$
$y ( max ) 0 ≈ 0.0943 τ 2 [ ( 1 + 0.260 τ 2 + 0.365 τ 4 ) - 0.121 τ ( 1 + 0.31 τ 2 + 0.14 τ 4 ) ] ,$
$Y T ( max ) 01 F T 2 ( max ) / Z m ≡ Θ ( max ) 01 ≈ 0.0218 τ 4 × [ ( 1 - 1.463 τ 2 + 0.722 τ 4 ) - 0.171 τ ( 1 - 1.51 τ 2 + 0.67 τ 4 ) ] .$
$Θ ( max ) 01 = 0.00082 ,$
$Y T ( max ) 01 = ( 0.00082 ) F T 2 ( max ) /$
$Θ ( max ) 01 = 0.00096 ,$
$Y T ( max ) 01 = 0.00096 F T 2 ( max ) / Z m .$
$[ Y T ( max ) 01 ] P = 0.21 mm ,$
$[ Y T ( max ) 01 ] N = 0.24 mm .$
$( F 1 ) 0 = + ( 1.400 ) ( 40.00 ) mm = + 56.00 mm , ( F 2 ) 0 = - ( 1.154 ) ( 40.00 ) mm = - 46.16 mm ,$
$( F 3 ) 01 = + ( 3.436 ) ( 40.00 ) mm = + 137.44 mm .$
$( F max ) 01 = + ( 5.440 ) ( 40.00 ) mm = + 217.60 mm ,$
$[ L 1 ′ ( 0 ) ] 0 = + ( 3.859 ) ( 40.00 ) mm = + 154.36 mm .$
$( F 1 ) 0 = - ( 1.149 ) ( 40.00 ) mm = - 45.96 mm , ( F 2 ) 0 = + ( 1.421 ) ( 40.00 ) mm = + 56.84 mm , ( F 3 ) 01 = - ( 2.955 ) ( 40.00 ) mm = - 118.20 mm , ( F max ) 01 = - ( 5.339 ) ( 40.00 ) mm = - 213.56 mm ,$
$[ L 1 ′ ( 0 ) ] 0 = - ( 2.380 ) ( 40.00 ) mm = - 95.20 mm .$
$( r ) P = ℛ = 3.0 , and ( τ ) P = + 0.5.$
$∊ = 0.080.$
$z 1 = 0 , z 2 = 0.420 , z 3 = 1.00.$
$γ 1 = 1.420 , γ 2 = 0.420 , γ 3 = 0 ,$
$m = 1 2 ( 1 / 2 τ + ∊ ) = 0.540.$
$f 3 = + 3.5205 , s 32 = 1.1500 , f 2 = - 1.1199 , s 21 = 0.1500 , f 1 = + 1.4413 , } or , with Z m = 40.0 mm , F 3 = + 140.82 mm , s 32 = 46.00 mm , F 2 = - 44.80 mm , s 21 = 6.00 mm , F 1 = + 57.65 mm .$
$L 1 ′ ( z ) ≈ ( 3.9404 - z ) 40.0 mm = ( 157.62 - 40.0 z ) mm ,$
$F ( z ) = 5.4661 1 + 1.0381 z + 0.9619 z 2 40.0 mm .$
$W = ( 5.4661 ) ( 40.0 ) / 100.0 = 2.1864.$
$Y T ( z ) = 250.0 [ 0.0322 z ( z - 0.420 ) ( z - 1.000 ) 1 + 1.0381 z + 0.9619 z 2 ] mm .$
$f 3 = - 2.8906 , s 32 = 1.1500 , f 2 = + 1.4605 , s 21 = 0.1500 , f 1 = - 1.1133 , } or , with Z m = 40.0 mm , F 3 = - 115.62 mm , S 32 = 46.00 mm , F 2 = + 58.60 mm , S 21 = 6.00 mm , F 1 = - 44.53 mm .$
$L 1 ′ ( z ) ≈ - [ 3.3106 - ( 1.0 - z ) ] 40.0 mm = - [ 132.42 - 40.0 ( 1.0 - z ) ] mm$
$F ( z ) = - [ 5.3051 / ( 3 - 3.1253 z + 1.1253 z 2 ) ] 40.0 mm .$
$W = - ( 5.3051 ) ( 40.0 ) / 100.0 = - 2.1220.$
$L = 172.06 mm = 1.72 F T ( max ) .$
$Y T ( z ) = 250.0 [ 0.0400 z ( z - 0.580 ) ( z - 1.000 ) 3 - 3.1253 z + 1.1253 z 2 ] mm .$
$R≜FmaxFmin=fmaxfmin.$
$r≜F(0)F(1)=f(0)f(1).$
$τ≜f(0)-f(1)f(0)+f(1)=r-1r+1=±R-1R+1,$