Download Exercise 13 Geometrical and Technical Optics WS 2013/2014

Exercise 13
Geometrical and Technical Optics WS 2013/2014
Achromatization with diffractive optical elements
In the lecture a hybrid achromatic lens consisting of a refractive lens and a diffractive optical
element has been discussed.
a) Design a hybrid achromatic lens with a planoconvex lens made of BK7 and a plane
diffractive optical element on the plane surface of the refractive lens. The achromatic lens
should have a focal length of f’=100 mm for the wavelength d=587.6 nm and it is
illuminated by a plane wave with lateral diameter of 40 mm. The center thickness of the
refractive lens should be 6 mm. First, the diffractive lens should only have a parabolic
phase function =2ar2 and a=-1/(2df’2) (focal length of the diffractive lens f’2).
b) Calculate the phase function of the diffractive lens in such a way that it corrects the
spherical aberration for the wavelength d. In order to do so, use the holographic principle,
i.e., a spherical wave going from the paraxial image point to the DOE plane defines the
image wave and the plane wave refracted at the spherical front surface and propagated to
the DOE plane defines the reference wave. To calculate the phase function as polynomial
(radial polynomial with only even orders) use the function “Phase Function” in the
evaluation menu.
c) Calculate the chromatic aberrations and the point aberrations for the hybrid achromatic
lens of b) for the wavelengths 486.1 nm, 587.6 nm and 656.3 nm. Compare the results
with that of a single planoconvex lens with focal length 100 mm made of BK7.
d) In exercise 11 we designed a pure refractive achromatic lens with also about 100 mm
focal length. The data (optimized for a beam diameter of 40 mm) are: R1=68.818 mm,
R2=-49.523 mm, R3=-107.738 mm. Material between the first two surfaces with distance
15 mm is BK7 and material between the second and third surface with distance 5 mm is
SF10. Compare that element with the hybrid achromatic lens of b).
Solution:
To a)
The focal lengths of the refractive lens f’1 and the diffractive lens f’2 have to fulfill the
condition:
f '1 V1, d   f '2 V2,d
The Abbe numbers are: V1,d=64.167 (BK7) and V2,d= -3.452
The total focal length of the system is (assuming thin lenses which are cemented):
1
1
1


f ' f '1 f '2

V  V2, d
V
1
1
1 V1, d  V2,d

 2,d 
 f '1  f ' 1,d
 105.38 mm
f ' f '1 V1, d f '1 f '1
V1, d
V1,d

1
1
 n  1  R  n  1 f '1  1.5168  1  105.38 mm  54.46 mm
f '1
R
And :
V
V  V1,d
1
1
1 V2,d  V1, d

 1,d 
 f '2  f ' 2,d
 1958.8 mm
f ' f '2 V2,d f '2 f '2
V2,d
V2, d
 a
1
2d f '2
 0.4344 mm - 2
To b)
Take first the setup of a) and remove the diffractive back surface. Then, trace the rays and
calculate the phase function in the DOE plane (ref).
Then, take the paraxial image point (for this take in a) a very small numerical aperture and
calculate the focus) at 96.43 mm behind the back surface as new point light source and trace
in free space up to the DOE plane. There, calculate again the phase function (im).
Finally, calculate the phase function of the DOE itself. This is:
 DOE   out   in   im   ref   im   ref 
Add the two phase functions in the menu “Tools –> Operations with phase functions” and
multiply them with a factor -1.
Then, the coefficients are:
a_2=-0.434424451808 mm-2
a_4=0.0004936895243 mm-4
a_6=-1.8680154e-008 mm-6
To c)
For the hybrid achromatic lens and lateral diameter of 40 mm of the illuminating wave it is
(first surface at z=-6 mm and second surface at z=0 mm):
=587.6 nm: z_Focus: 96.431 mm, spot radius: 1.58e-5 mm, P/V wave aberrations: 0.0008 
=486.1 nm: z_Focus: 95.740 mm, spot radius: 0.0260 mm, P/V wave aberrations: 3.918 
=656.3 nm: z_Focus: 96.657 mm, spot radius: 0.0174 mm, P/V wave aberrations: 1.917 
Taking a paraxial ray bundle (lateral diameter of incident wave was set to only 0.04 mm), the
axial focus positions are at:
=587.6 nm: z_Focus: 96.431 mm
=486.1 nm: z_Focus: 96.235 mm
=656.3 nm: z_Focus: 96.326 mm
So, the maximum axial distance difference is about 0.2 mm.
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A pure planoconvex lens with 100 mm focal length at d has a radius of curvature of:
R  n  1 f '  51.68 mm
In this case the aberrations for 40 mm beam diameter are (again first surface at z=-6 mm and
second surface at z=0 mm):
=587.6 nm: z_Focus: 92.981 mm, spot radius: 0.161 mm, P/V wave aberrations: 20.57 
=486.1 nm: z_Focus: 91.955 mm, spot radius: 0.162 mm, P/V wave aberrations: 25.18 
=656.3 nm: z_Focus: 93.444 mm, spot radius: 0.161 mm, P/V wave aberrations: 19.32 
The axial focus positions for only 0.04 mm beam diameter are:
=587.6 nm: z_Focus: 96.045 mm
=486.1 nm: z_Focus: 94.991 mm
=656.3 nm: z_Focus: 96.520 mm
So, the maximum axial distance difference is about 1.5 mm.
To d)
The absolute positions of the focus depend of course on the selected z-positions of the
surfaces. If we set the first surface to z=-17 mm it is (beam diameter 40 mm):
=587.6 nm: z_Focus: 96.087 mm, spot radius: 0.0316 mm, P/V wave aberrations: 4.956 
=486.1 nm: z_Focus: 96.242 mm, spot radius: 0.0226 mm, P/V wave aberrations: 4.704 
=656.3 nm: z_Focus: 96.099 mm, spot radius: 0.0349 mm, P/V wave aberrations: 4.786 
The axial focus positions for only 0.04 mm beam diameter are:
=587.6 nm: z_Focus: 96.940 mm
=486.1 nm: z_Focus: 96.924 mm
=656.3 nm: z_Focus: 97.017 mm
So, the maximum axial distance difference is about 0.09 mm.
Summarizing, the refractive achromatic lens has lower chromatic errors than the hybrid
achromatic lens (by about a factor 2 lower). On the other side, the spherical aberration of the
refractive achromatic lens is higher, especially for d.
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