Съдържание

Условие Отговори Решение

OPhO 2026, Единствен кръг — Задача 9. Beam Splitter

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Всички материали от OPhO 2026

2026 Online Physics Olympiad · 21 август 2026 г.

Last updated

Last updated: 22nd August 03:35 UTC

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Corrections (all times are in UTC)

Q8: Cleared up wording (21/08 04:40 and 04:52)
Q3: improved wording (21/08 08:49)
Q19: Improved wording (21/08 09:42)
Q16: Attempts reset (21/08 22:00), Error bound increased (22/8 02:30)
Q1: Improved wording (22/08 03:35)

Указания

Format. The OPhO is a team-based competition (3 members at most per team). Unlike previous years, the 2026 OPhO consists of a single round held from August 21, 12 am (UTC), to August 23, 11:59 pm (UTC). The contest is made up of three parts, all running concurrently over the full contest window:

Short Answer — problems requiring numerical answers, submitted on the OPhO portal;
Long Answer — problems requiring full written solutions, submitted on the OPhO portal;
Simulation — computational problems requiring teams to run a provided executable locally, with solutions submitted on the OPhO portal.

All teams have access to all three parts for the entire duration of the contest. Submissions are made on the OPhO portal https://opho.physoly.org/ using the team login, and may be made at any time within the contest window.

Short Answer. Teams submit numerical answers on the portal, with up to three attempts per problem. Every answer submission must be accompanied by a file upload, on the portal, of the working used to obtain that answer. This working will not be graded, but submissions without it will not be accepted. The base score for each problem depends on the number of teams that successfully answer it, each incorrect attempt multiplies the base score by a factor of 0.8, and there is no time factor. The score for an individual problem follows:

w(N) = (0.8)^i\left[e^{-\frac{x(N)}{200}} + e^{-\frac{x(N)^2}{100}} + 1\right],

where $i \in \{0, 1, 2\}$ represents whether a team got it on the first, second, or third try, $N$ represents the number of teams that answer the problem correctly, and $x(N)$ is the sum of $(0.8)^i$ across all teams that have solved the problem. The base score is independent of problem number, but expect questions to be ordered roughly by difficulty. A team’s raw Short Answer total is the sum of these values, and all raw totals are then scaled linearly such that the top team receives 55 points.

Long Answer. Teams submit full solutions on the portal. Rubrics are predetermined, and solutions will be reviewed to determine the proper number of points to give. The points breakdown is given per problem, for a total of 35 points. It is recommended that participants write their solutions in LaTeX. Handwritten solutions (or a combination of both) are accepted too; if participants have more than one photo of a handwritten solution (jpg, png, etc.), it is required that they be organized in the correct order in a pdf before submitting.

Simulation. Teams will need to be able to run a provided executable locally. Solutions are submitted on the portal and will be graded against predetermined rubrics. The points breakdown is given per problem, for a total of 10 points.

Scoring. A team’s cumulative score is the sum of its scaled Short Answer score (out of 55 points), Long Answer score (out of 35 points), and Simulation score (out of 10 points), for a maximum score of 100 points.

Permitted Resources. Use of the internet is allowed, including textbooks, published papers, calculators, and computer algebra systems. However, the use of AI tools of any kind is strictly prohibited. This includes, but is not limited to, large language models (e.g. ChatGPT, Claude, Gemini), AI coding assistants, AI-generated search summaries, and any other AI-powered assistance.

Dishonest Behavior. Please refrain from cheating in any form. This includes, but is not limited to, creating alternate accounts to gain additional attempts, sharing answers with other teams, posting problems online, utilizing AI tools as described above, or employing any other method to gain an unfair advantage.

We have previously caught numerous teams engaging in such practices, and pursuing dishonest tactics provides no genuine sense of security. Any teams found to be acting maliciously will be notified via email, along with the opportunity to appeal. Failing to provide a valid appeal will result in a ban from participating in any future Physoly competitions. Our utmost priority is to ensure an enjoyable contest experience for all participants, and we remain committed to preventing any disruption to the fun for others.

Awards and Results. The top competitors will be recognized on the website. No cash prizes will be awarded this year.

List of Constants

Proton mass, $m_p = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Neutron mass, $m_n = 1.67 \cdot 10^{-27}\ \mathrm{kg}$
Electron mass, $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$
Avogadro’s constant, $N_0 = 6.02 \cdot 10^{23}\ \mathrm{mol^{-1}}$
Universal gas constant, $R = 8.31\ \mathrm{J/(mol \cdot K)}$
Boltzmann’s constant, $k_B = 1.38 \cdot 10^{-23}\ \mathrm{J/K}$
Electron charge magnitude, $e = 1.60 \cdot 10^{-19}\ \mathrm{C}$
1 electron volt, $1\ \mathrm{eV} = 1.60 \cdot 10^{-19}\ \mathrm{J}$
Speed of light, $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$
Universal Gravitational constant, $G = 6.67 \cdot 10^{-11}\ \mathrm{(N \cdot m^2)/kg^2}$
Solar Mass $M_\odot = 1.988 \cdot 10^{30}\ \mathrm{kg}$
Acceleration due to gravity, $g = 9.8\ \mathrm{m/s^2}$
1 unified atomic mass unit, $1\ \mathrm{u} = 1.66 \cdot 10^{-27}\ \mathrm{kg} = 931\ \mathrm{MeV/c^2}$
Planck’s constant, $h = 6.63 \cdot 10^{-34}\ \mathrm{J \cdot s} = 4.41 \cdot 10^{-15}\ \mathrm{eV \cdot s}$
Permittivity of free space, $\epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C^2/(N \cdot m^2)}$
Coulomb’s law constant, $k = \frac{1}{4\pi\epsilon_0} = 8.99 \cdot 10^{9}\ \mathrm{(N \cdot m^2)/C^2}$
Permeability of free space, $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{T \cdot m/A}$
Magnetic constant, $\frac{\mu_0}{4\pi} = 1 \cdot 10^{-7}\ \mathrm{(T \cdot m)/A}$
1 atmospheric pressure, $1\ \mathrm{atm} = 1.01 \cdot 10^{5}\ \mathrm{N/m^2} = 1.01 \cdot 10^{5}\ \mathrm{Pa}$
Wien’s displacement constant, $b = 2.9 \cdot 10^{-3}\ \mathrm{m \cdot K}$
Stefan-Boltzmann constant, $\sigma = 5.67 \cdot 10^{-8}\ \mathrm{W/m^2/K^4}$

Условие

Zeus, an aspiring optics enthusiast, fires a laser producing a beam of unpolarized light. Being somewhat careless with expensive equipment, he manages to trap this beam inside a glass slab of refractive index $n = 4$ , length $l$ , and breadth $b = 2\,\mathrm{m}$ , with the beam propagating at an angle $\theta = 30^{\circ}$ to the normal. Zeus positions a camera at the far end inside the slab.

The camera can distinguish between the two linearly polarized components if the separation between the centers of the two beams is $x = 1\,\mathrm{mm}$ . In order for the two components to be distinguishable, what is the minimum length $\ell$ (in meters) of the glass slab? The light has a vacuum wavelength $\lambda = 633\,\mathrm{nm}$ , and the slab is surrounded by air.

Neglect the diffraction effects of the laser.

Отговори

Решение

Let a plane, $s$ -polarised wave travel in a medium of index $n_1$ and be incident on an interface with a medium of lower index $n_2 < n_1$ , at an angle $\theta_1$ exceeding the critical angle $\theta_c = \arcsin(n_2/n_1)$ . Choose the $x$ -axis along the interface and the $y$ -axis along its normal, and write the incident field as

E_i = E_0\exp\big[i(\omega t - \vec{k} \cdot \vec{r})\big], \qquad \vec{k} \cdot \vec{r} = k\sin\theta\,x + k\cos\theta\,y,

with $k = 2\pi/\lambda_1$ and $\lambda_1 = \lambda/n_1$ the wavelength in the incident medium.
The Fresnel coefficient for $s$ -polarisation is

r_s = \frac{n_1\cos\theta_1 - n_2\cos\theta_2}{n_1\cos\theta_1 + n_2\cos\theta_2}.

Beyond the critical angle Snell’s law yields a complex transmitted angle, and

n_2\cos\theta_2 = n_2\sqrt{1 - \frac{n_1^2}{n_2^2}\sin^2\theta_1} = i\sqrt{n_1^2\sin^2\theta_1 - n_2^2},

so that $r_s$ is the ratio of a complex number and its conjugate. Hence $|r_s| = 1$ and

E_r = E_i\exp(2i\phi_0), \qquad \phi_0(\theta_1) = \arctan\left(\frac{1}{\cos\theta_1}\sqrt{\sin^2\theta_1 - \frac{n_2^2}{n_1^2}}\right).

The reflection is total, and its sole effect is a phase shift $2\phi_0$ that depends on the angle of incidence.
A physical beam is not a single plane wave but a packet of wavevectors spanning a narrow range of angles about a central value $\theta_0$ . Writing the angular spectrum as $A(\theta)$ and suppressing the common factor $e^{i\omega t}$ ,

E_{\mathrm{in}} = \int A(\theta)\,\exp\big[-i\,(k\sin\theta\,x + k\cos\theta\,y)\big]\,d\theta,

E_{\mathrm{out}} = \int A(\theta)\,\exp\big[2i\phi_0(\theta)\big]\exp\big[-i\,(k\sin\theta\,x + k\cos\theta\,y)\big]\,d\theta,

each component acquiring its own phase shift on reflection.
Since the spectrum is narrow, expand to first order in $(\theta - \theta_0)$ :

\phi_0(\theta) \simeq \phi_0(\theta_0) + (\theta - \theta_0)\frac{\partial\phi_0}{\partial\theta}, \qquad \sin\theta \simeq \sin\theta_0 + (\theta - \theta_0)\cos\theta_0, \qquad \cos\theta \simeq \cos\theta_0 - (\theta - \theta_0)\sin\theta_0.

Substituting, the two fields separate into a carrier wave times a slowly varying envelope:

E_{\mathrm{in}} = e^{-i(k\sin\theta_0 x + k\cos\theta_0 y)}\int A(\theta)\,\exp\big[-i(\theta - \theta_0)(k\cos\theta_0\,x - k\sin\theta_0\,y)\big]\,d\theta,

E_{\mathrm{out}} = e^{2i\phi_0}\,e^{-i(k\sin\theta_0 x + k\cos\theta_0 y)}\int A(\theta)\,\exp\left[i(\theta - \theta_0)\left(2\frac{\partial\phi_0}{\partial\theta} - k\cos\theta_0\,x + k\sin\theta_0\,y\right)\right]d\theta.

The extra term $2(\theta - \theta_0)\,\partial_\theta\phi_0$ is precisely what would be produced by translating the argument $x$ of the envelope, so that

\boxed{E_{\mathrm{out}}(x, y) = e^{2i\phi_0}\,E_{\mathrm{in}}\left(x - \frac{2}{k\cos\theta_0}\frac{\partial\phi_0}{\partial\theta},\ y\right)}

That is, the reflected beam emerges displaced along the interface by $\Delta x = \dfrac{2}{k\cos\theta_0}\dfrac{\partial\phi_0}{\partial\theta}$ , rather than from the geometrical point of incidence.
Differentiating $\phi_0$ , with $n \equiv n_1/n_2$ ,

\frac{\partial\phi_0}{\partial\theta} = \frac{\sin\theta}{\sqrt{\sin^2\theta - n^{-2}}}.

The displacement measured perpendicular to the beam axis is $D_s = \Delta x\cos\theta_0$ , whence

D_s = \frac{2}{k}\frac{\partial\phi_0}{\partial\theta} = \frac{\lambda_1}{\pi}\frac{n\sin\theta}{\sqrt{n^2\sin^2\theta - 1}}, \qquad \lambda_1 = \frac{\lambda}{n_1}.

The shift is of the order of a wavelength, and diverges as $\theta \to \theta_c$ , where the first-order expansion (and hence this treatment) ceases to be valid.
This result gives the lateral displacement $D_s$ for s-polarized light. An entirely analogous calculation for p-polarized light yields

D_p = \frac{D_s}{\sin^2\theta\,(n^2 + 1) - 1}.

For the two polarization components to be distinguished by the camera, the accumulated difference in lateral displacement after $N$ total internal reflections must equal the prescribed separation $x$ :

N\,|D_s - D_p| = x.

Since each reflection corresponds to a longitudinal advance of $b\tan\theta$ along the slab, the minimum slab length required is

l_{\min} = N\,b\tan\theta = \frac{b\,x\,\tan\theta}{|D_s - D_p|}.

Substituting the given values yields

\boxed{l_{\min} = \frac{26\pi bx}{9\lambda} = 28675\ \mathrm{m}}.

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf