Achromat Lens Design

Design Goal

Design an achromatic doublet with a sodium D focal length of 50 mm using the N-BAF10 crown and SF10 flint glasses. Constrain the interior surface that is mutual to both singlets to have zero curvature.

Initial Calculation

First and foremost, we need to determine the focal lengths of the crown and flint singlets.

From the definition of Abbe V-number, VV

ΦF−ΦCΦD=1VD\frac{\Phi_F - \Phi_C}{\Phi_{D}} = \frac{1}{V_D}

ΦF−ΦC=ΦDVD\Phi_F - \Phi_C = \frac{\Phi_{D}}{V_D}

where
ΦF\Phi_F is the power of the lens at the sodium-F line (blue)

ΦC\Phi_C is the power of the lens at the sodium-C line (red)

ΦD\Phi_D is the power of the lens at the sodium-d line (yellow)

VDV_D is the Abbe V-number for the material at the sodium-d line (yellow)

Considering a doublet,

ΦD,Doublet=ΦD,1+ΦD,2\Phi_{D,Doublet} = \Phi_{D,1} + \Phi_{D,2}

Individual lens, i.e. Lens 1 and Lens 2 has the following dispersion equations:

ΦF,1−ΦC,1=ΦD,1V1\Phi_{F,1} - \Phi_{C,1} = \frac{\Phi_{D,1}}{V_1} -------- (1)

ΦF,2−ΦC,2=ΦD,2V2\Phi_{F,2} - \Phi_{C,2} = \frac{\Phi_{D,2}}{V_2} -------- (2)

By summing up equation (1) and (2),

(ΦF,1+ΦF,2)−(ΦC,1+ΦC,2)=ΦD,1V1+ΦD,2V2(\Phi_{F,1} + \Phi_{F,2}) - (\Phi_{C,1} + \Phi_{C,2}) = \frac{\Phi_{D,1}}{V_1} + \frac{\Phi_{D,2}}{V_2}

ΦF,Doublet−ΦC,Doublet=ΦD,1V1+ΦD,2V2\Phi_{F,Doublet} - \Phi_{C,Doublet} = \frac{\Phi_{D,1}}{V_1} + \frac{\Phi_{D,2}}{V_2}

To design a achromatic doublet means to set the dispersion, i.e. ΦF,Doublet−ΦC,Doublet\Phi_{F,Doublet} - \Phi_{C,Doublet} to zero, hence

0=ΦD,1V1+ΦD,2V20 = \frac{\Phi_{D,1}}{V_1} + \frac{\Phi_{D,2}}{V_2}

ΦD,1V1=−ΦD,2V2\frac{\Phi_{D,1}}{V_1} = - \frac{\Phi_{D,2}}{V_2}

ΦD,1=−V1V2ΦD,2\Phi_{D,1}=-\frac{V_1}{V_2}\Phi_{D,2}



From the fact the lens power adds up, for sodium-d line:

ΦD=ΦD,1+ΦD,2\Phi_{D}=\Phi_{D,1}+\Phi_{D,2}
ΦD,2=ΦD−ΦD,1\Phi_{D,2}=\Phi_{D}-\Phi_{D,1}


ΦD,1=−V1V2ΦD,2\Phi_{D,1}=-\frac{V_1}{V_2}\Phi_{D,2}
ΦD,1=−V1V2(ΦD−ΦD,1)\Phi_{D,1}=-\frac{V_1}{V_2}(\Phi_{D}-\Phi_{D,1})
ΦD,1(1−V1V2)=−V1V2ΦD\Phi_{D,1}(1-\frac{V_1}{V_2})=-\frac{V_1}{V_2}\Phi_{D}
ΦD,1(V2−V1V2)=−V1V2ΦD\Phi_{D,1}(\frac{V_2-V_1}{V_2})=-\frac{V_1}{V_2}\Phi_{D}
ΦD,1=V1V1−V2ΦD\Phi_{D,1}=\frac{V_1}{V_1-V_2}\Phi_{D}

From OpticStudio's Material Catalog, we can obtain the Abbe V-number for the crown and flint glass, i.e. 47.112 and 28.41 respectively.

By plugging in the Abbe V-numbers,

ΦD,1=V1V1−V2ΦD\Phi_{D,1}=\frac{V_1}{V_1-V_2}\Phi_{D}
ΦD,1=47.11247.112−28.41(150)\Phi_{D,1}=\frac{47.112}{47.112-28.41}(\frac{1}{50})
ΦD,1=0.0504mm−1\Phi_{D,1}=0.0504 mm^{-1}
fD,1=1ΦD,1=19.85mmf_{D,1} = \frac{1}{\Phi_{D,1}}=19.85 mm

ΦD,2=ΦD−ΦD,1\Phi_{D,2}=\Phi_{D}-\Phi_{D,1}
ΦD,2=150−119.85\Phi_{D,2}=\frac{1}{50}-\frac{1}{19.85}
ΦD,2=−0.03038mm−1\Phi_{D,2}=-0.03038 mm^{-1}
fD,2=1ΦD,2=−32.91mmf_{D,2} = \frac{1}{\Phi_{D,2}} = -32.91 mm

Next, we need to determine the radius of curvature for surface 1 and surface 3 (considering interior surface, i.e. surface 2, that is mutual to both singlets is constrained to have zero curvature)

Φ=(n−1)(c1−c2+(n−1)dc1c2n)\Phi = (n-1)(c_1-c_2+\frac{(n-1)dc_1c_2}{n})

Given c2=0c_2 = 0,

ΦD,1=(nD,1−1)(c1)\Phi_{D,1} = (n_{D,1} - 1)(c_1)
0.0504=(1.67−1)c10.0504 = (1.67 - 1)c_1

0.0504=0.67c10.0504 = 0.67c_1
c1=0.05040.67=0.0752c_1 = \frac{0.0504}{0.67} = 0.0752
r1=1c1=10.0752=13.294r_1 = \frac{1}{c_1} = \frac{1}{0.0752} = 13.294 mm


ΦD,2=(nD,2−1)(0−c3)\Phi_{D,2} = (n_{D,2} - 1)(0 -c_3)
−0.03038=(1.7283−1)(−c3)-0.03038 = (1.7283 - 1)(-c_3)
−0.03038=−0.7283c3-0.03038 = -0.7283c_3
c3=−0.03038−0.7283=0.0417c_3 = \frac{-0.03038}{-0.7283} = 0.0417
r3=1c3=10.0417=23.973r_3 = \frac{1}{c_3} = \frac{1}{0.0417} = 23.973 mm

OpticStudio Layout

To approximate the design condition of thin lenses, let the on-axis thickness of the crown be 2 mm and the on-axis thickness of the flint be 1 mm.

The field point at 0°0\degree from infinity is set to have a Clear Semi-Dia of 4.5 mm. The lens' Clear Semi-Dia is set to be 4.7 mm, and the Aperture Type was chosen to be Entrance Pupil Diameter with value > 4.5 mm.

The image plane distance is set to solve for marginal ray height equal 0.

The Number of Rays setting in Layout was set to be 3 so that only the chief ray and 2 marginal rays were displayed.

The final lens layout is as shown below:

Zoomed-in illustration of the image plane:

The image formed by the marginal rays appears in front of the paraxial image plane. A further zoomed-in illustration is as shown below:

The cyan rays represent blue wavelength, the orange rays represent yellow wavelength, and the red rays represent the red wavelength.

Interpreting the Longitudinal Aberration plot

The Longitudinal Abberation view shows the offset due to longitudinal chromatic aberration.

The title of the plot shows Pupil Radius, which is equal to the radius of the effective entrance pupil. Since the limiting aperture in our lens layout is the Clear Semi-Dia of the field point, hence the Pupil Radius shown here is 4.5 mm.

The Normalized Pupil Coordinate axis of the plot here indicates the position that the object space ray (from infinity) enters the entrance pupil.

The value 1.0 means the ray is marginal ray that happens to be the edge of the entrance pupil (or a point at the top of the bundle of rays entering the system).

The value 0 means the ray is equivalent to a point at the center of the ray bundle.

The change in effective focal length FC−FFF_C-F_F​ of this achromat lens (on-axis, i.e. Rel-Pupil= 0) is

FC−FF=9.6×10−2−(−1.6×10−1)F_C-F_F = 9.6 \times 10^{-2} - (-1.6 \times 10^{-1})
FC−FF=0.096+0.16=0.256F_C-F_F = 0.096 + 0.16 = 0.256 mm