total time: 125min Herbert Gross Ralf Hambach Friedrich Schiller University Jena Institute of Applied Physics Albert-Einstein-Str 15 07745 Jena

Imaging and Aberration Theory – Exam

Task 1: Primary Aberrations ( 7 / 70 ) 10min

(a) Name the 5 Seidel aberrations and 2 main chromatic aberrations for a rotationally symmetric optical system imaging an object plane to an image plane.

(b) Consider a system made of thin lenses. Which aberrations are influenced by

the diameter of the stop,

the location of the stop, if the NA of the system is kept constant,

a bending of the thin lenses?

aberration influenced by

stop size stop shift bending

Solution:

aberration stop size stop shift bending

spherical x x

Coma x x x

Astigmatism x x x

Petzval Field Curvature

Distortion x x

Lateral Color x x

Axial Color

Petzval and axial color only depend on focal power distribution (of a system of thin

lenses): 1

𝑛′𝑅𝑃𝑇𝑍= − ∑

𝐹𝑘

𝑛𝑘𝑘 , Δ𝑠𝐶𝐻𝐿 = −

𝑠′2

𝑤2 ∑ 𝑤𝑘2 𝐹𝑘

𝜈𝑘𝑘 , 𝑤𝑘 … relative marginal ray height.

stop size: chief ray is not changed, marginal ray changes

stop shift: chief ray is changed, i.e. ray heights at single lenses changes. Spherical only depends on NA, which is kept constant.

bending: affects incidence angles at surfaces, changes position of principle planes

Points:

(a) 2 Pts. for all names (one missing: 1 Pt.)

(b)

o 2 Pts. Petzval and axial color not influenced

o 1 Pt. stop size: Sph., Coma, Ast. correct

o 1 Pt. stop shift: ~Sph, Coma,Ast,Dist., Lateral Color correct

o 1 Pt. bending: Sph, Coma, Ast. correct

Task 2: Perfect Imaging ( 9 / 70 ) 15min

(a) Consider the focusing of a collimated beam. Name two single-element systems, which

produce a perfect geometric spot (indicate the orientation if necessary). Which fundamental physical principle has to be fulfilled in the case of perfect imaging?

(b) Consider the (virtual) imaging of an on-axis point object P to its image P’. Name two single-element systems, which produce a perfect image point P’ (indicate the orientation if necessary). Which of the two systems is better for the imaging of a small, extended object?

(c) Derive a condition for perfect imaging of a small, extended object in the plane perpendicular to the optical axis. Hint: Start from the variation of the point-point Eikonal due to a small deviation from the conjugate points P and P’,

(d) Briefly discuss, if an object volume can be imaged perfectly.

Solution: (a) grin-lens, parabolic mirror, elliptical surface between two media, planar-hyperbolic

asphere The Fermat principle has to be fulfilled, i.e. the optical path length of all rays to the focal point has to be the same.

(b) grin-lens, elliptical mirror, aplanatic-concentric lens (virtual image). The aplanatic surface fulfills the sine-condition, i.e. linear coma vanishes and extended objects are imaged perfectly (see (c)).

(c) The point-point Eikonal 𝑉(𝑟, 𝑟′) for the two conjugate on-axis points P and P’ is

independent of the ray angles 𝑠 and 𝑠′. If we consider the point-point Eikonal for off-axis points Q and Q’ in close proximity to P and P’, we can approximate in first order of the

deviation PQ= 𝑑𝑟, P’Q’= 𝑑𝑟′: 𝑉(𝑟 + 𝑑𝑟, 𝑟′ + 𝑑𝑟′) ≈ 𝑉(𝑟, 𝑟′) − 𝑛 𝑠 ⋅ 𝑑𝑟 + 𝑛′𝑠′ ⋅ 𝑑𝑟′. If Q and Q’ shall be also conjugate, then this Eikonal must be also independent of the ray angles 𝑠

and 𝑠′, i.e.

𝑑𝑉 = −𝑛 𝑠 ⋅ 𝑑𝑟 + 𝑛′𝑠′ ⋅ 𝑑𝑟′ = Const(𝑑𝑟, 𝑑𝑟′) = 0.

For points Q and Q’ in the transverse plane, 𝑑𝑟, 𝑑𝑟′ ⊥ 𝑒𝑧, the constant must be 0 as we can see by choosing the special ray 𝑠 = 𝑠′ = 𝑒𝑧 along the optical axis. For arbitrary rays

with incident ray direction 𝑠, we thus obtain

𝑛 𝑠 ⋅ 𝑑𝑟 = 𝑛′𝑠′ ⋅ 𝑑𝑟′, 𝑛 sin 𝑢 𝑑𝑟 = 𝑛′ sin 𝑢′ 𝑑𝑟′.

(Sine condition)

(d) No, if we consider homogenous object and image space. Yes, if we use GRIN medium, like Maxwell fisheye lens.

Points:

(a) 2 Pts. (one for each correct system, if uniquely specified)

1 Pt. Fermat principle + equal OPL

(b) 2 Pts. (one for each correct system, if uniquely specified)

1 Pt. aplanatic system

(c) 1 Pt. Ansatz

1 Pt. solution (sine condition), also without derivation

(d) 1 Pt. correct explanation of yes or no. (+1 for both explanations)

Task 3: Paraxial Raytrace ( 10 / 70 ) 15min

In paraxial approximation, any rotationally symmetric optical system may be expressed by an ABCD matrix relating the height y and angle u of a ray at given entrance and exit planes.

(a) How many degrees of freedom does a paraxial optical system possess? Why?

(b) Show that any paraxial optical system can be represented by a system of two thin lenses.

(c) Derive an expression of the focal length and the back-focal length of the system in terms of the ABCD parameters.

(d) Consider the reversed system (-z direction). Under which condition is the focal length and when is the back-focal length unchanged compared to the original system?

Solution:

(a) 3, the 4 parameters ABCD are coupled via the Lagrange invariant: 𝐴𝐷 − 𝐵𝐶 = 𝑛/𝑛′.

(b) two lenses with distance d in between have the ABCD matrix

( 1 0

−𝐹1 1) (

1 𝑑/𝑛0 1

) ( 1 0

−𝐹2 1) = (

1 − 𝑑𝐹2/𝑛 𝑑/𝑛−𝐹1 − 𝐹2 + 𝑑𝐹1𝐹2/𝑛 1 − 𝑑𝐹1/𝑛

)

hence, d is fixed by B, 𝐹1 by B and D, and 𝐹2 by A and B.

(c) Ansatz: incoming parallel ray: [𝑦′

𝑢′] = (

𝐴 𝐵𝐶 𝐷

) [𝑦𝑢

], [𝐴𝐶

] = ( 𝐴 𝐵𝐶 𝐷

) [10

]

focal length: measured from principle plane: 1

𝑓′ = −𝑢′, 𝑓′ = −1

𝐶.

back-focal length: measured from vertex: 𝑦′

𝐵𝐹𝐿= −𝑢′, 𝐵𝐹𝐿 = −

𝐴

𝐶

(d) Ansatz: reversed system is like tracing rays backwards, i.e. calculating (y,u) from (y’,u’).

The system matrix is given by the inverse of the ABCD matrix:

[𝑦𝑢

] =1

𝐴𝐷 − 𝐵𝐶(

𝐷 −𝐵−𝐶 𝐴

) [𝑦𝑢

] =𝑛′

𝑛(

𝐷 −𝐵−𝐶 𝐴

) [𝑦𝑢

]

focal length: 𝑓′ =𝑛

𝑛′

1

𝐶 is unchanged (except for a sign) if 𝑛 = 𝑛′

back-focal length: 𝐵𝐹𝐿 =𝐷

𝐶 is unchanged (except for sign) if 𝐴 = 𝐷.

Points:

Note: both notations for ABCD matrices (using (y,u) and (y,nu) as vectors) are correct.

(a) 1 Pt. correct number

1 Pt. Lagrange invariant

(b) 2 Pts

(c) 1 Pt. Ansatz

1 Pt. f

1 Pt. BFL

(d) 1 Pt. Ansatz

1 Pt. n=n’ for same f

1 Pt. A=D for same BFL

Task 4: Spot shape of Secondary Coma ( 10 / 70 ) 15min The wave aberration of secondary coma of 5th order reads in polar coordinates on a

normalized pupil radius 𝑊(𝑦, 𝜌, 𝜙) = 𝐴5𝑦𝜌5𝑐𝑜𝑠𝜙, where 𝐴5 is the peak value of the aberration.

(a) Show that the transverse ray errors Δ𝑥′, Δ𝑦′ in the image plane located at a distance R

from the exit pupil are given by Δ𝑥′ ∝ −4𝜌4 sin 𝜙 cos 𝜙, Δ𝑦′ ∝ −𝜌4[4 cos2 𝜙 + 1].

(b) Sketch the tangential and sagittal ray fan plots for this aberration.

(c) Sketch the pattern of the image spot of a point object (contour lines for ray cones with fixed exit pupil radius 𝜌). Indicate the tangential and sagittal coma on the sketch.

(d) Determine the centroid position of the spot relative to the chief ray.

Hint: sin 2𝜙 = 2 sin 𝜙 cos 𝜙, cos 2𝜙 = cos2 𝜙 − sin2 𝜙 = 2 cos2 𝜙 − 1.

Solution:

(a) Ansatz: Δ𝑥′ ≈ −𝑅

𝑛′

𝑑

𝑑𝑥𝑝𝑊(𝑥𝑝, 𝑦𝑝) = −

𝑅

𝑛′ 𝐴5𝑦 𝑑

𝑑𝑥𝑝(𝑥𝑝

2 + 𝑦𝑝2)

2 𝑦𝑝 and analogous for Δ𝑦′.

we obtain Δ𝑥′ = −𝑅

𝑛′ 𝐴5𝑦 (𝑥𝑝2 + 𝑦𝑝

2) 4𝑥𝑝𝑦𝑝 ∝ −𝜌4 2 sin 2𝜙

Δ𝑦′ = −𝑅

𝑛′ 𝐴5𝑦 (𝑥𝑝2 + 𝑦𝑝

2) (4𝑦𝑝2 + (𝑥𝑝

2 + 𝑦𝑝2)) ∝ −𝜌4[2 cos 2𝜙 + 3]

(b) tangential: Δ𝑦′(𝑥𝑝 = 0, 𝑦𝑝) ∝ −5𝑦𝑝4

sagittal: Δ𝑥′(𝑥𝑝, 𝑦𝑝 = 0) = 0

(c) same as coma spot, different apex angle.

tangential point (𝑥𝑝 = 0, 𝑦𝑝): Δ𝑥′ = 0, Δ𝑦′ = −5𝑦𝑝4

sagittal point (𝑥𝑝, 𝑦𝑝 = 0): Δ𝑥′ = 0, Δ𝑦′ = −𝑥𝑝4

(d) centroid for fixed 𝜌 is the center of the circle, offset by Δ𝑦′(𝜌) ∝ −3𝜌4. Averaging over all 𝜌 gives

⟨Δ𝑦′⟩ = ∫ 𝜌𝑑𝜌1

0

∫ 𝑑𝜙2𝜋

0

Δ𝑦′(𝜌, 𝜙) = ∫ 𝜌𝑑𝜌1

0

(−3)𝜌4 = −3

6[𝜌6]0

1 = −1

2

Points:

(a) 1 Pt. Ansatz

2 Pts. derivation x,y

(b) 2 Pts

(c) 1 Pt. Sketch

1 Pt. tangential point

1 Pt. sagittal point

(d) 1 Pt. Ansatz

1 Pt. result

tangential

sagittal

Task 5: Focusing of collimated beam ( 7 / 70 ) 10min

Consider the focusing of a collimated, monochromatic beam by a single, plano-convex lens.

(a) Which orientation of the lens is preferred for a minimal spot diameter at the focus? Why?

(b) How does the spot diameter vary, if an additional stop at the lens is changed in size? Consider diffraction and aberration effects.

(c) Sketch the focal size as a function of the stop diameter.

(d) Name three approaches to reduce the minimal achievable spot diameter.

Solution: a) Curved surface first. Distributes refraction over two surfaces, which reduces spherical aberration. b) If the diameter is small, the geometrical aberrations are negligible and the spot is blurred due to

diffraction according to the Airy formula D = /NA. If the diameter is large, the geometrical aberrations causes a large blur compared to the diffraction effects. c)

Dspot

NA / D

diffraction

Airy

geometrical

aberration

total

d) Possible answers:

lens bending (to reduce Spherical)

make aspherical surface (“)

more complicated optical system: classical achromate to reduce Spherical

Points:

(a) 1 Pt. correct orientation + reason

(b) 2 Pts

(c) 1 Pt.

(d) 3 Pt.

Task 6: Dyson System ( 8 / 70 ) 20min The so called Dyson system is a very clever setup, which can be used for some imaging applications with a quite good correction. A typical layout is shown in the following figure. All curved surfaces are spherical, the surface in the object and image plane respectively is plane.

mirror

objecty

imagey'

rL

n

rM

tL

ta

(a) What are the remaining primary aberrations of the system in the general case of arbitrary parameters? Why? Sketch the unfolded system, where the mirror is taken as thin lens.

(b) If the parameters are chosen as 𝑡𝐿 = −𝑟𝐿 and 𝑡𝐿 + 𝑡𝑎 = −𝑟𝑀, what is the consequence for the remaining aberrations?

(c) Give the condition for correcting Petzval field curvature in 3rd order. Calculate the radius 𝑟𝐿 of the lens, if the glass index is n = 1.6 and the mirror has a radius 𝑟𝑀 = −160𝑚𝑚.

Solution: a) The unfolded system is seen here, it is completely symmetric. Therefore the aberrations: 1. coma 2. lateral color 3. distortion are perfectly corrected.

b) If the condition tL = -rL and tL+ta = -rM is fulfilled, the spherical mirror and the curved lens surface are used concentric for the object point on axis. Therefore spherical aberration and axial chromatical aberration is perfectly corrected.

c) The lens has the focal length

1

n

rf L

L

The mirror has

2

MM

Rf

The Petzval theorem states a vanishing 3rd order field curvature, if the condition

j jj fn

01

is fulfilled. In the case of the Dyson system, this corresponds to the condition

LM

M

LML

rr

rn

nr

n

rnr

n

0121

for the choice of the refractive index. We get the radius of the lens as mmrn

nr ML 60

1

Points:

(a) 1 Pt. symmetry, odd aberrations vanish

1 Pt. coma, lateral color, distortion vanish

1 Pt. sketch (positive mirror, positive lenses, symmetric setup)

(b) 1 Pt. concentric → spherical = 0

1 Pt. axial color = 0

(c) 1 Pt. Ansatz

1 Pt. correct signs and derivation

1 Pt. results

Task 7: Zernike vs. Seidel Expansion ( 5 / 70 ) 10min Briefly explain the Zernike- and Seidel approach to describe wave aberrations in an optical system. Give three differences between the individual approaches. Solution: Seidel

Seidel compares the perturbation of the real ray to the paraxial one expands the deviations on the base of the Eikonal into a Taylor series in fourth order. The individual prefactors can be identified as the summed surface contributions of the individual third order primary aberrations.

The main advantage is the access to the individual surface contribution of the primary aberrations.

The main disadvantage is its restriction as a third order expansion - It is only valid for negligible influence of higher order aberrations - present higher order aberrations can lead to misinterpretations.

Zernike

The Zernike description relies on the comparison of the real wavefront with an ideal reference wavefront in the exit pupil. The wavefront error is decomposed into the contributions of individual aberrations, on the basis of the orthogonal Zernike polynomials.

The Zernike expansion is not restricted to third order aberrations.

Zernikes are only given for one field point and one wavelength

The Zernike expansion is appropriate for the evaluation of aberrations on experimental grounds with interferometric methods.

Points:

2 Pts. definition

3 Pts. differences

Task 8: Aspherical Achromate ( 8 / 70 ) 15 min

Consider a classical achromate (cemented doublet) for focusing a 10mm wide, collimated beam with an image-sided numerical aperture of 0.1. The thickness of the lens is neglected.

(a) Determine the focal length of the first and second lens, which are made of BK7 and SF12.

(b) Fixing the two focal lengths, the classical achromate with spherical surfaces has one degree of freedom left. Which aberration is further corrected in the classical achromate?

(c) Briefly explain the remaining aberrations of a classical achromate: residual zonal error, secondary spectrum, and spherochromatism. Indicate these errors in the longitudinal aberration diagram below.

(d) Which of the remaining aberrations can be reduced by making one of the surfaces aspheric? Which aberration cannot be influenced by making even all surfaces aspheric?

Solution:

(a) effective focal length of lens: 𝑓 ≈𝐷

2𝑁𝐴= 50𝑚𝑚, 𝐹 = 0.02𝑚𝑚−1

Condition for the focal power: 21 FFF (1)

Condition for achromatization 02

2

1

1

FF (2)

Therefore by solving for the individual focal powers (in air):

mmF

fmmmmF

F 63.231

,04232.0116.202.0

1 1

1

11

1

2

1

(3)

mmF

fmmF

F 81.441

,02232.0116.102.0

1 2

2

1

2

1

2

(b) bending the lens → spherical aberration can be corrected (in some cases)

(c) residual zonal error: residual aberration for green light at intermediate pupil values

secondary spectrum: remaining axial color for red and blue

Glass 𝒏𝒅 Abbe Number

BK7 1.5168 64.167336

SF12 1.6483 33.841187

spherochromatism: difference in spherical aberration for red and blue

(d) Making the first surface aspherical, on can achieve perfect imaging for one wavelength

(green). Making front and rear suface aspherical, one can also correct

spherochromatism. Axial color cannot be influenced, as it is fixed by the paraxial values.

Points:

(a) 1 Pt. Ansatz

1 Pt. result

(b) 1 Pt spherical

(c) 3 Pts (one per aberration, correct explanation + sketch)

(d) 1 Pt. correct green over entire pupil

1 Pt. axial color remains

Task 9: MTF of a Pinhole Camera ( 6 / 70 ) 15min Consider a pinhole camera shown in the figure below. Assume that the object is incoherent,

monochromatic, and very far away 𝑧0 ≫ 𝑧𝑖. The circular pinhole has a radius 𝑎.

(a) Calculate the incoherent modulation transfer function of the camera for the case of a large pinhole, such that the point spread function can be estimated using geometrical optics. What is the cutoff frequency of the system? (Hint: Find the intensity point-spread function, then Fourier transform it.).

(b) Calculate the cutoff frequency for a sufficiently small pinhole diameter, such that diffraction effects dominate.

(c) Discuss the optimal size of the pinhole qualitatively.

Solution: (a) The PSF is given by geometrical shadow of pinhole created by point-like source.

𝐼𝑃𝑆𝐹(𝑟) = circ(𝑟/𝑎′), 𝑎′ =𝑧𝑜+𝑧𝑖

𝑧𝑜𝑎 ≈ 𝑎.

The incoherent MTF is the Fourier trafo if the PSF, i.e., an Airy pattern

𝐻𝑀𝑇𝐹(𝑘) = ℱ{𝐼𝑃𝑆𝐹}(𝑘, 0) = 2𝐽1(2𝜋𝑘𝑎)

2𝜋𝑘𝑎, 𝑘𝑚𝑎𝑥 =

0.622

𝑎,

which has a first zero at 𝑘𝑚𝑎𝑥. After 𝑘𝑚𝑎𝑥, contrast is reversed (until next zero occurs).

(b) Diffraction occurs, when the pinhole is very small. In this case, the PSF corresponds to the intensity of the (Fraunhofer) diffraction pattern of a homogeneously illuminated

circular aperture of radius 𝑎. One has

𝐼𝑃𝑆𝐹(𝑟) = [2𝐽1(2𝜋𝑟NA′/𝜆 )

2𝜋𝑟NA′/𝜆]

2, with the first zero at 𝑟𝐴𝑖𝑟𝑦 =

0.622 𝜆

NA′ =0.622 𝜆𝑧𝑖

𝑎.

The inhoherent MTF is given by the Fourier trafo of the PSF intensity, i.e. by the auto-convolution of the Fourier trafo of the Airy disk. This is the same as the convolution of two circles with radius

𝑘0 =𝑁𝐴′

𝜆=

𝑎

𝜆𝑧𝑖, which results in 𝑘𝑚𝑎𝑥 = 2𝑘0 =

2𝑎

𝜆𝑧𝑖.

(c)

Dspot

NA / D

diffraction

Airy

geometrical

aberration

total

Points:

(a) 1 Pt. disk-like PSF

1 Pt. Airy pattern

1 Pt. kmax

(b) 1 Pt. Airy disk as PSF (ideal transfer function)

1 Pt. kmax (even without calculation)

(c) 1 Pt. Sketch or discussion: optimum exists