Балкански 2026, Теория — Задача 1. Phase-detection autofocus

Concours Général de Physique «Minko Balkanski» · 31 юли 2026 г. · 10 т.

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Условие

This problem studies a simplified phase-detection autofocus system of the type used in single-lens reflex cameras. The main camera objective is modelled as a thin converging lens (L)(L) with optical centre OO, object and image foci FF and F′F', and image focal length f′=OF′‾>0f' = \overline{OF'} > 0.

After unfolding the optical path produced by two redirecting mirrors, the autofocus module can be represented on the same axis as the objective. It contains a field lens (LC)(LC) of focal length fc′=12 mmf'_c = 12\ \mathrm{mm}, two separator microlenses (LS1)(LS_1) and (LS2)(LS_2) of equal focal length fs′=3 mmf'_s = 3\ \mathrm{mm}, and two linear sensors (C1)(C_1) and (C2)(C_2). The microlens centres are separated by 2a=2 mm2a = 2\ \mathrm{mm}. Each sensor contains 48 equally spaced photodiodes along a 1 mm1\ \mathrm{mm} line; use the sampling pitch d=1 mm/48d = 1\ \mathrm{mm}/48.

An electronic unit compares the two recorded intensity profiles. Their relative shift gives the sign and magnitude of the focusing error; a controller then drives a servomotor that translates the focusing group.

I. Formation of a sharp image

1. State the Gaussian conditions and approximations under which the thin lens (L)(L) is used.

2. The object is represented by a segment (AB)(AB) perpendicular to the optical axis and its image by (A′B′)(A'B'). The point CC is where the image sensor meets the axis. Define

p=OA‾,p′=OA′‾,x=CA‾,x0=CO‾,x′=CA′‾.p = \overline{OA},\qquad p' = \overline{OA'},\qquad x = \overline{CA},\qquad x_0 = \overline{CO},\qquad x' = \overline{CA'}.

Algebraic distances are positive from left to right, in the direction of light propagation; transverse segments are positive upwards and angles are positive counterclockwise.

Optical axis with tick marks; object AB drawn as a vertical arrow with B at the top and A on the axis, labelled x below; focus F on the axis; the thin lens (L) as a vertical double arrow with centre O and x0 below; image focus F'; the sensor at point C on the right; a horizontal double arrow below labelled D spanning from the object to the sensor; at the right, sign conventions with + arrows upwards, to the right and a counterclockwise angle.
Figure I.1 – Image construction by the camera lens.

2. a) Starting from the Descartes thin-lens equation

1−p+1p′=1f′,\frac{1}{-p} + \frac{1}{p'} = \frac{1}{f'},

derive the transverse magnification relation connecting AB‾\overline{AB}, A′B′‾\overline{A'B'}, pp and p′p', and hence obtain Newton's conjugation relation connecting FA‾\overline{FA}, F′A′‾\overline{F'A'} and f′f'.

2. b) On the copy of the figure provided on the supplementary sheet, construct (A′B′)(A'B') with a ruler using two principal rays. Check the position of A′A' using Newton's relation. Is the image formed on the sensor?

3. Let D=∣AC‾∣D = |\overline{AC}| be the object-to-sensor distance and T=OC‾=−x0>0T = \overline{OC} = -x_0 > 0 the lens-to-sensor distance.

3. a) Assuming that the objective lies between the object and sensor, obtain the two possible solutions for TT. Retain the solution T<D/2T < D/2, corresponding to the objective close to the sensor, and show that

T=D−D2−4Df′2.T = \frac{D - \sqrt{D^2 - 4Df'}}{2}.

3. b) Find the condition on DD and f′f' for these positions to be real. Give a geometrical interpretation of the limiting case.

3. c) Determine T∞T_\infty when the object is at infinity.

3. d) For f′=50 mmf' = 50\ \mathrm{mm}, the objective must focus between infinity and Dmin=40 cmD_{\mathrm{min}} = 40\ \mathrm{cm}. Calculate the required travel δ=T(Dmin)−T∞\delta = T(D_{\mathrm{min}}) - T_\infty. In which direction must the lens move as the object approaches?

3. e) A full-frame sensor of dimensions 36 mm×24 mm36\ \mathrm{mm} \times 24\ \mathrm{mm} contains 24 million square pixels. The camera must resolve the 10 cm10\ \mathrm{cm} width of an object at 2.0 km2.0\ \mathrm{km}; require its image to span at least two pixels. Neglecting diffraction and aberrations, determine the minimum focal length. The numerical data in this part are independent of part d).

II. Study of the autofocus module

After unfolding the optical path, the system is represented below. The point SS is where the microlens plane meets the axis and PP is where the photosensor plane meets it.

When focus is correct, the intermediate image (A′B′)(A'B') is formed in the plane of the field lens (LC)(LC), whose optical centre is OcO_c. The microlenses lie in the image focal plane of (LC)(LC), so OcS=fc′O_cS = f'_c.

Unfolded optical axis: the objective (L) with centre O at the left; at distance T the field lens (LC) with centre Oc, the intermediate image A'B' (A' on the axis, B' below); dashed chief rays at angles β1 and β2 going to the two microlenses (LS1) and (LS2) with centres Os1 and Os2, each at distance a from the axis; the sensors C1 and C2 on the right; distances f'c from (LC) to the microlens plane and SP from the microlenses to the sensors; y and x axes at the right.
Figure I.2 – Unfolded model of the phase-detection autofocus module.

4. First consider the correctly focused configuration.

4. a) The plane containing (LC)(LC) is conjugate to the photosensor plane through each microlens. Determine the distance SPSP.

4. b) On the corresponding figure on the supplementary sheet, complete the construction of the secondary images (A1′′B1′′)(A''_1B''_1) and (A2′′B2′′)(A''_2B''_2). The points A1′′A''_1 and A2′′A''_2, which are images of the axial point A′A', lie at the respective centres of (C1)(C_1) and (C2)(C_2). Compare the orientation of the secondary images with that of (A′B′)(A'B').

4. c) Explain the role of the field lens: why does it allow both microlenses to receive bundles from the same point of the intermediate image? Which measured quantity carries the defocus information?

4. d) For the axial point A′A', use the geometry of the figure and the small-angle approximation to determine the two chief-ray angles β1\beta_1 and β2\beta_2 in terms of aa and fc′f'_c. State their signs, then calculate their magnitude.

5. The objective remains fixed and the initially focused object moves by a small distance Δx>0\Delta x > 0 towards the camera. The primary image point then moves by Δx′\Delta x' beyond the sensor plane. Use f′=50 mmf' = 50\ \mathrm{mm} for numerical applications.

5. a) Starting from the thin-lens equation, show to first order that

Δx′=(TD−T)2Δx.\Delta x' = \left(\frac{T}{D - T}\right)^2 \Delta x.

For D=1.0 mD = 1.0\ \mathrm{m} and Δx=1.0 cm\Delta x = 1.0\ \mathrm{cm}, first calculate Δx′\Delta x' using the exact value of TT obtained in Question 3a), then give the estimate obtained from D≫TD \gg T and T≃f′T \simeq f'.

5. b) A′A', which lies to the right of (LC)(LC), is a virtual object for that lens. If Ac′A'_c is its image through (LC)(LC), show that

OcAc′‾=fc′ Δx′fc′+Δx′.\overline{O_cA'_c} = \frac{f'_c\,\Delta x'}{f'_c + \Delta x'}.

Justify the approximation OcAc′‾≃Δx′\overline{O_cA'_c} \simeq \Delta x'.

5. c) ⋆\star Using the approximation from part b), let s=fc′−Δx′s = f'_c - \Delta x' be the new object distance for the microlenses. Use their conjugation relation to show that both secondary images undergo the same longitudinal displacement

Δx1′′=Δx2′′=(fs′fc′−fs′)2Δx′.\Delta x''_1 = \Delta x''_2 = \left(\frac{f'_s}{f'_c - f'_s}\right)^2 \Delta x'.

5. d) ⋆\star Using the rays through the microlens centres, show to first order that

Δβ1=+a Δx′fc′2,Δβ2=−a Δx′fc′2.\Delta\beta_1 = +\frac{a\,\Delta x'}{f'^2_c},\qquad \Delta\beta_2 = -\frac{a\,\Delta x'}{f'^2_c}.

In the fixed photosensor plane, the centre of each profile follows the chief ray. Using the distance SPSP found in Question 4a), show that

Δy1′′=SP Δβ1=+afs′fc′(fc′−fs′)Δx′,Δy2′′=SP Δβ2=−afs′fc′(fc′−fs′)Δx′.\Delta y''_1 = SP\,\Delta\beta_1 = +\frac{a f'_s}{f'_c(f'_c - f'_s)}\Delta x',\qquad \Delta y''_2 = SP\,\Delta\beta_2 = -\frac{a f'_s}{f'_c(f'_c - f'_s)}\Delta x'.

Calculate Δx1′′\Delta x''_1, Δx2′′\Delta x''_2, Δy1′′\Delta y''_1 and Δy2′′\Delta y''_2 numerically.

5. e) Describe the relative shift of the two intensity profiles on the sensors and calculate its value. Compare it with the photodiode pitch dd. Briefly explain why correlation of the complete intensity profiles may nevertheless measure a shift smaller than one pitch. How does the sign of the shift distinguish focus in front of the sensor from focus behind it?

6. The objective is now moved so that the image is formed again on the sensor. The algebraic lens displacement ΔxL\Delta x_L is positive in the direction of light propagation.

6. a) What is the sign of ΔxL\Delta x_L when the object has approached, so that Δx>0\Delta x > 0?

6. b) Linearising the thin-lens equation while the object and lens move simultaneously, show that

ΔxL=KLΔx,KL=−T2D(D−2T).\Delta x_L = K_L \Delta x,\qquad K_L = -\frac{T^2}{D(D - 2T)}.

For D=1.0 mD = 1.0\ \mathrm{m} and Δx=1.0 cm\Delta x = 1.0\ \mathrm{cm}, first calculate KLK_L and ΔxL\Delta x_L using the exact value of TT, then give the estimate obtained from D≫TD \gg T and T≃f′T \simeq f'.

7. Describe a complete autofocus loop: acquisition of the two profiles, estimation of their relative shift, determination of the error sign, motor command, and stopping criterion. Also give one situation in which phase-detection autofocus performs poorly.

Решение

Внимание: Непълно решение

В архива няма официално решение на тази задача.

Annexe : constantes et relations utiles — Reference sheet : constants and useful relations

Constantes communes / Common constants

G=6.67×10−11 m3 kg−1 s−2,σ=5.67×10−8 W m−2 K−4,G = 6.67 \times 10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}},\qquad \sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}},
1 an=1 yr=3.16×107 s,ln⁡2=0.693,1 rad=57.3∘.1\ \mathrm{an} = 1\ \mathrm{yr} = 3.16 \times 10^{7}\ \mathrm{s},\qquad \ln 2 = 0.693,\qquad 1\ \mathrm{rad} = 57.3^{\circ}.

Français

Les relations ci-dessous peuvent être utilisées directement. Elles ne remplacent pas le choix et la justification du modèle physique.

Problème I – petites variations et optique

Pour ∣θ∣≪1 rad|\theta| \ll 1\ \mathrm{rad} :

sin⁡θ≃tan⁡θ≃θ.\sin\theta \simeq \tan\theta \simeq \theta.

Au premier ordre :

F(z+Δz)−F(z)≃F′(z)Δz,F(z + \Delta z) - F(z) \simeq F'(z)\Delta z,
Δ(1z)≃−Δzz2.\Delta\left(\frac{1}{z}\right) \simeq -\frac{\Delta z}{z^2}.

Problème II – seuil, équations et intégrales

Pour la question 3d) :

qc=c(Ts−T0)+xcL,q_c = c(T_s - T_0) + x_c L,
y=2qcχu∞2,μ(1+μ)2=y,y = \frac{2q_c}{\chi u_\infty^2},\qquad \frac{\mu}{(1+\mu)^2} = y,
yμ2+(2y−1)μ+y=0.y\mu^2 + (2y - 1)\mu + y = 0.

La racine physique vérifie 0<μ<10 < \mu < 1.

Si dv/dt=(vt−v)/τdv/dt = (v_t - v)/\tau et v(0)=0v(0) = 0 :

v(t)=vt(1−e−t/τ).v(t) = v_t\left(1 - e^{-t/\tau}\right).

Si dr/dt=−krdr/dt = -kr :

ln⁡(r2/r1)=−k(t2−t1).\ln(r_2/r_1) = -k(t_2 - t_1).

Intégrales utiles :

∫r/2rs ds=3r28,∫0Rr4 dr=R55,\int_{r/2}^{r} s\,ds = \frac{3r^2}{8},\qquad \int_0^R r^4\,dr = \frac{R^5}{5},
∫T−4 dT=−13T3.\int T^{-4}\,dT = -\frac{1}{3T^3}.

Pour ∣z∣≪1|z| \ll 1 : (1−z)3≃1−3z(1 - z)^3 \simeq 1 - 3z.

English

The relations below may be used directly. They do not replace the choice and justification of a physical model.

Problem I – small changes and optics

For ∣θ∣≪1 rad|\theta| \ll 1\ \mathrm{rad}:

sin⁡θ≃tan⁡θ≃θ.\sin\theta \simeq \tan\theta \simeq \theta.

To first order:

F(z+Δz)−F(z)≃F′(z)Δz,F(z + \Delta z) - F(z) \simeq F'(z)\Delta z,
Δ(1z)≃−Δzz2.\Delta\left(\frac{1}{z}\right) \simeq -\frac{\Delta z}{z^2}.

Problem II – threshold, equations, and integrals

For Question 3d):

qc=c(Ts−T0)+xcL,q_c = c(T_s - T_0) + x_c L,
y=2qcχu∞2,μ(1+μ)2=y,y = \frac{2q_c}{\chi u_\infty^2},\qquad \frac{\mu}{(1+\mu)^2} = y,
yμ2+(2y−1)μ+y=0.y\mu^2 + (2y - 1)\mu + y = 0.

The physical root satisfies 0<μ<10 < \mu < 1.

If dv/dt=(vt−v)/τdv/dt = (v_t - v)/\tau and v(0)=0v(0) = 0:

v(t)=vt(1−e−t/τ).v(t) = v_t\left(1 - e^{-t/\tau}\right).

If dr/dt=−krdr/dt = -kr:

ln⁡(r2/r1)=−k(t2−t1).\ln(r_2/r_1) = -k(t_2 - t_1).

Useful integrals:

∫r/2rs ds=3r28,∫0Rr4 dr=R55,\int_{r/2}^{r} s\,ds = \frac{3r^2}{8},\qquad \int_0^R r^4\,dr = \frac{R^5}{5},
∫T−4 dT=−13T3.\int T^{-4}\,dT = -\frac{1}{3T^3}.

For ∣z∣≪1|z| \ll 1: (1−z)3≃1−3z(1 - z)^3 \simeq 1 - 3z.


Оригинал в Архива: CGP_2026.pdf