Съдържание

Условие Optical Tweezers Acoustic Levitation Решение

OPhO 2021, Теория — Задача 1. Levitation

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от OPhO 2021

2021 Online Physics Olympiad: Invitational Contest — v1.1 Theoretical Exam · 25 т.

Условие

Levitation is a widely researched area in physics with wide applications in real life. Commercial high speed trains use magnetic levitation to transport passengers and the very physics of aerodynamics helps planes fly in the sky. Although the many uses of levitation physics apply on macroscopic dimensions, this problem will also analyze the various applications of levitation on the microscopic scale.

Optical Tweezers

An optical tweezer (OT) is a device which uses tightly focused laser beams to trap an object in all three spacial dimensions $(x-y-z)$ . Consider using an OT to trap a dielectric polystyrene nanosphere with mass $m$ , radius $R$ , and relative dielectric constant $\varepsilon_r$ . A laser beam is directed to the vertical $z$ -direction (the laser beam is similar to a monochromatic light wave propagating a sparse medium.) The laser beam has a wavelength $\lambda$ . The time averaged intensity in the $z$ -direction due to light can be considered to follow a Gaussian distribution function:

I(\rho,z)=I_0\left(\frac{W_0}{W(z)}\right)^2\exp\left(\frac{-2\rho^2}{W(z)^2}\right)

where $\rho$ is the distance from the center of the beam and $W_0$ is known as the waist size, or the measure of the beam size at the point of its focus. Here the waist length, in general, follows $W(z)=W_0\sqrt{1+z^2/z_R^2}$ where $z_R=\pi W_0^2/\lambda$ denotes the Rayleigh length.

Complete nanosphere displaced to the left of the x/z axes inside the Gaussian beam, with beam outlines and upward intensity arrows. — Figure 1: A nanosphere placed off center in a Gaussian beam.

Native colored Gaussian intensity heatmap with complete coordinate ticks, color bar and W0 double arrow. — Figure 2: A graphic of the intensity distribution in a Gaussian beam.

The OT traps particles via three different forces: scattering forces created by the change in momentum of light scattered or absorbed by a particle; gradient forces due to the polarization of the particle created by the strong electric fields of the laser beam; and, radiation forces produced by an accelerating charge. The total power radiated by an oscillating electric dipole with dipole moment $p_0$ at frequency $\omega$ will be $P_R=\frac{\mu_0p_0^2\omega^4}{12\pi c}$ , where $c$ is the speed of light.

1. (6 pts.) In the Rayleigh spectrum, the nanosphere’s size is such that $\lambda\gg R$ . Find the oscillation frequency $\Omega$ and equilibrium position of the nanosphere when slightly displaced a distance $d\ll W_0$ in the $x$ -direction. Neglect the scattering forces in this part.

In the Mie spectrum, the particle size is no longer negligible such as in part A, and follows $R\gtrsim\lambda$ . As a result, a non-homogenous electric field is incident upon the sphere and scattering forces are no longer neglectable. Parts B and C will investigate the nanosphere in the Mie spectrum.

1. The order of magnitude of $R$ is greater than $\lambda$ .

2. (5 pts.) Determine the scattering force and torque on the nanosphere as a function of a distance $x\ll W_0$ from its origin. We can consider $x\sim R$ . Assume that $100\%$ of light is transmitted for simplicity. The index of refraction of the nanosphere is $n$ while the index of refraction of the medium is $m$ . You may express your answer as an integral if needed.

2. In reality, some light will be reflected due to Fresnel’s equations.

3. (2 pts.) In a simplified model, the torque acting on the nanosphere can be represented as $\boldsymbol\tau=\kappa\boldsymbol\omega$ where $\kappa$ is a numerical constant and $\boldsymbol\omega$ is the angular velocity of the nanosphere. At $t=0$ , the nanosphere is stationary with an angular velocity $\omega_0$ and the OT is turned on. After a time $T$ , the OT is turned off with the nanosphere left floating in the medium again. Determine the angular velocity of the nanosphere after a time $t$ passes where $t$ is the time of the entire process. It can be expressed as a piecewise function.

Acoustic Levitation

Objects can also be trapped via sound waves. Consider the simplest way to model sound waves; that is, in one dimension. A cylindrical tube of length $L_0$ with ambient temperature and pressure $T_0$ and $P_0$ is fixed with a piston at one end of cross-sectional area $S$ that moves periodically as $x(t)=A\cos(2\pi ft)$ where $A\ll L_0$ is the amplitude of the piston and $f$ is the frequency of the process. The tube contains $n$ monatomic particles of mass $m$ per unit volume. As the piston moves back and forth, the air in the tube compresses or expands (rarefies) in the tube as a sound wave. The molecules within the tube move back or forth parallel to their equilibrium position as the air travels within the piston. Neglect any viscous or turbulent friction within the pipe.

3. In some models, the piston can represent the moving cone of a loud speaker.

4. (6 pts.) What is the average power required to move the piston? Consider the limits of $f\gg c_s/A$ and $f\ll c_s/A$ where $c_s$ is the speed of the sound wave.

In sound waves, the density perturbations are very small, so it can be assumed that $\Delta\rho\ll\rho_0$ . where $\rho_0$ is the original density of the pipe. Furthermore, the wavelength of the sound waves are much larger than the mean free path of the gas molecules. The sound wave created by the oscillating piston moves at a speed $c_s$ and dissipates the power given by the piston.

4. You cannot use $\Delta\rho$ as a variable in this problem, but you can use $\rho_0$ .

5. (4 pts.) As a result of compression, the air within the sound wave has a larger temperature $\Delta T\ll T_0$ . Find the change in temperature $\Delta T$ . If you were unable to solve problem 4, express the average power to move the piston as $P$ .

6. (2 pts.) Consider a small cylindrical object of radius $R<\sqrt S$ and width $h\ll R$ in the pipe where variations of pressure on the cylinder’s surface are negligible. Determine the force $F$ acting on the cylinder when the sound wave passes through it. If the pipe is placed on a vertical plane where gravity is present, qualitatively describe what location(s) the cylinder would levitate.

Complete closed acoustic tube, oscillating blue piston with red double arrow, density shading and full L0 dimension arrow. — Figure 3: A visualization of how acoustic waves within the pipe are created via the oscillation of the piston.

Published organizer clarifications (separate source, pages 1–2)

Quoted from the organizer clarification document:

Part 1:

For problem 1, neglect scattering forces completely in all subparts. Radiation forces are not negligible.
When finding the scattering force, the variation in $z$ is neglectable.
For question 2, you may assume that $x\ll W_0$ and $x\sim R$ .
The origin is the center of the optical tweezer.
You may assume all photons are directed in $+z$ direction.
~~The angular velocity of the nanosphere at $t=0$ is 0 in question 3.~~ EDIT: The following was wrong. You must consider the angular velocity of the nanosphere at $t=0$ to be $\omega_0$ .
The torque obtained from a simplified model is not the general solution to part 2. You do not need to add contributions from part 2 to this part. If the torque is positive, it works to increase omega. Furthermore, the time $t$ described is asking during the entire process and not just after the turnoff, so you may express your answer as a piecewise function.
Torque is about the center of the sphere.

Part 2:

You only need to find upper and lower limits in question 4 and not general.
The pipe does not conduct heat with its surroundings and is closed. The power that is put in the piston is dissipated by the wave. You may assume the rest of the energy from the wave is absorbed by the end of the pipe once the sound wave collides with it (reflection coefficient is 0).
Air within the sound wave refers to the average temperature within the maximally compressed regions of the wave.
The answer for problem 4 does not need to be expressed in terms of the speed of sound. The gas is monatomic and ideal.
Delta rho is not a variable that you can use in this problem. However, you can use rho.
You have to consider a cylinder in question 6.
In question 6, there is no need to write a differential equation for when the pipe is vertical. We only ask for you to qualitatively describe where the cylinder is located and not quantitatively. The wave is moving so you can describe its general position as there will be multiple equilibrium locations and the cylinder will be oscillating.

Решение

Внимание: Непълно решение

The complete available printed key is retained. For question3, it gives the exponential spin law but no explicit turn-off branch or complete piecewise expression involving T.

1. When the exam was administered, it was said that to neglect the scattering forces in this part. In reality, the radiation forces are a type of scattering force as it takes light and re-emits it. For ambiguity, the equilibrium position was not considered in grading.

Optical Tweezers — (a)

The Rayleigh approximation tells us that $\lambda\gg R$ . Thus, the electric field around the microsphere is homogenous; hence, the polarization of the microsphere will be as well. Furthermore, since the nanoball is displaced a small distance $d\ll W_0$ , the intensity distribution can be approximated by Taylor expanding $e^x\approx1+x$ . In other words:

I(x)\approx I_0\left(1-\frac{2x^2}{W_0^2}\right).

Intensity and electric field can be related via the time averaged Poynting vector as as

I=\lvert\langle\vec S\rangle\rvert=\frac1{\mu_0}\lvert\langle\vec E\times\vec B\rangle\rvert=\frac1{c\mu_0}\frac{E_m^2}2

where $E_m$ is the maximum electric field of the Gaussian beam. As the nanoball becomes polarized, it obtains a dipole moment of $\vec p=-\alpha\vec E$ . It is required to find the value of $\alpha$ of a dielectric ball. The polarized sphere can be divided into two different polarized rods of charge densities $+\rho$ and $-\rho$ separated a distance $a$ , replacing them by patches of charge (circles) that are superimposed. Positive charges started accumulating on the upper half of the sphere while negative charges start accumulating on the lower half of the sphere as the electric field redistributes the charges.

Both native polarization illustrations kept together: field-line deformation around gray sphere and superimposed positive/negative density disks with full rho labels, r1/r2 and a vector arrows.

Thus, the field outside the small sphere is the sum of $E$ and the field due to an electric dipole that is at the center of the nanoball. The dipole moment expressed in terms of its polarization is

\vec p=\sum_iq\vec r_i=\vec P\cdot V=\frac43\pi R^3\vec P.

The total field at some point outside of the sphere is no longer uniform in the neighborhood of the sphere and is now the sum of the field $E$ and the field generated through the polarized matter, i.e. $\vec E=\vec E_0+\vec E_{\mathrm{dipole}}$ . Thus, the field depends on the polarization vector of the dielectric which can be expressed in terms of its electric susceptibility as

\vec P=\chi_e{}_0\vec E_{\mathrm{in}}=({}_r-1){}_0\vec E_{\mathrm{in}}.

Note that the potential at all the points on the spherical boundary is simply the dipole potential or

=\frac{p\cos\theta}{4\pi{}_0R^2}=\frac{PR\cos\theta}{3{}_0}.

Substituting $R\cos\theta=z$ implies $=Pz/3{}_0$ . By uniqueness theorem, this must be the only solution which means that $E_z=\frac\partial{\partial z}=-P/3{}_0$ . Hence, the internal field of the sphere follows $\vec E_{\mathrm{in}}=-\vec P/3{}_0$ . Superposition yields

\vec E_{\mathrm{in}}=\vec E_0-\frac{{}_r-1}3\vec E_{\mathrm{in}}\quad\Longrightarrow\quad\vec E_{\mathrm{in}}=\frac3{2+{}_r}\vec E.

Inserting this field into the dipole momentum tells us that

\vec p=\frac43\pi R^3({}_r-1){}_0\frac3{2+{}_r}\vec E\quad\Longrightarrow\quad\alpha=\frac{4\pi R^3({}_r-1){}_0}{2+{}_r}.

The gradient force on the ball depends on the gradients of the various components of its fields. In other words, $\vec F(t)=(\vec p\cdot\nabla)\vec E(t)$ . Averaging this quantity over time then allows to write

\langle\vec F(t)\rangle=\langle(\vec p\cdot\nabla)\vec E(t)\rangle=\langle(\alpha\vec E(t)\cdot\nabla)\vec E(t)\rangle=\frac\alpha2\nabla\langle E(t)^2\rangle=\frac\alpha2\nabla\frac{E_m^2}2=\frac{\alpha c\mu_0}2\partial_xI(x).

The standard differential equation for SHO can be retrieved from this as

m\ddot x=-\frac{2\alpha c\mu_0I_0}{W_0^2}x=-\kappa_xx\quad\Longrightarrow\quad\boxed{\Omega=\sqrt{\frac{2\alpha c\mu_0I_0}{W_0^2}}}.

Here, $\kappa_x$ is collectively known as the trap stiffness of the OT in the $x$ -direction. To find the force due to radiation in the $z$ -direction (as photons are directed on this axis), one can use the formula given in the problem for the power of radiation due to a rotating dipole. As $P=\vec F\cdot\vec v$ , one can write that $F=P/c$ . Thus,

F=\frac{\mu_0\alpha^2E^2\omega^4}{12\pi c^2}=\frac{\mu_0^2\alpha^2(2\pi c/\lambda)^4}{6\pi c}I(z)=\frac{8\pi^3c^3\mu_0^2\alpha^2}{3\lambda^4}I_0\left(1-\frac{2z^2}{W_0^2}\right).

This must be equal to the gradient force under equilibrium, thus creating a quadratic equation on $x$ :

F_{\mathrm{rad}}=F_{\mathrm{grad}}\quad\Longrightarrow\quad\frac{8\pi^3c^3\mu_0^2\alpha^2}{3\lambda^4}I_0\left(1-\frac{2z^2}{W_0^2}\right)=-\frac{2\alpha c\mu_0I_0}{W_0^2}z.

Substituting $\zeta\equiv\frac{4\pi^3c^2\mu_0\alpha W_0^2}{3\lambda^4}$ yields a quadratic of

\frac{2\zeta}{W_0^2}z^2-z-\zeta=0\quad\Longrightarrow\quad\boxed{z=\frac{W_0^2}{4\zeta}\left(1-\sqrt{1-\frac{8\zeta^2}{W_0^2}}\right)}.

6. This is well derived from the Poynting vector.

(b)

In the Mie spectrum, $\lambda\gg R$ which implies that the intensity and electric field distribution over the sphere is no longer homogenous. Create a ray diagram:

Complete refraction ray diagram inside circle: O, theta, alpha, beta, 2beta, kappa, all dashed radii and incoming/outgoing blue rays and vertical-axis arrow.

The deviation in angle can be described by the angle $\kappa$ or

\kappa=2(\alpha-\beta)=2\left(\alpha-\arcsin\left(\frac mn\sin\alpha\right)\right).

The change in momentum of photons that move towards the sphere can then be written as

\Delta p=\frac Ec(1-\cos\kappa)\quad\Longrightarrow\quad\mathrm dF=\frac{\Delta p}{\Delta t}=2I\sin^2\left(\alpha-\arcsin\left(\frac mn\sin\alpha\right)\right)\mathrm dA.

We can use the law of cosines for a point away from the origin to find $\rho$ :

\rho^2=x^2+(R\sin\theta)^2-2xR\sin\theta\cos\,.

Thus, putting everything together:

F_z=\frac{2I_0R^2}c\int_0^{\pi/2}\int_0^{2\pi}\sin^2\left(\alpha-\arcsin\left(\frac mn\sin\alpha\right)\right)\exp\left(\frac{-2\rho^2}{W_0^2}\right)\sin\theta\cos\theta\,\mathrm d\theta\mathrm d.

Similarly, you can write the forces in the $\hat x$ direction which will be similar to $F_z$ :

F_x=\frac{I_0R^2}c\int_0^{\pi/2}\int_0^{2\pi}\sin\left(\alpha-\arcsin\left(\frac mn\sin\alpha\right)\right)\exp\left(\frac{-2\rho^2}{W_0^2}\right)\sin\theta\cos\theta\cos\,\mathrm d\theta\mathrm d.

The net force is thus $F=\sqrt{F_x^2+F_z^2}$ . The torque acting on the sphere can be written as a volume integral over the cross product of distance and force or:

\tau=\int_V\vec r\times\vec F.

Here $\vec r$ will have different lengths along each direction, or $\vec r=(R\sin\alpha\cos\,,R\sin\alpha\sin\,,R\cos\alpha)$ .

(c)

The nanosphere has an angular momentum $L=I\omega$ which implies the torque acting on it to be

\tau=\frac{\mathrm dL}{\mathrm dt}=I\dot\omega=\kappa\omega.

We now have a simple first order differential equation whose solution is

\omega'=\omega_0\exp\left(\frac{5\kappa}{2mR^2}t\right).

As the nanosphere is an induced dipole, will it now lose power due to radiative losses? The answer turns out to be no, as the nanosphere rotates on an axis that is perpendicular to the direction of its dipole moment. With no dipole moment directed along the axis of rotation, no radiative power will be lost.

Grading Scheme

A1	It’s stated (or taken under consideration) that the entire field on the microsphere is uniform and explanation how oscillations occur is presented	0.50
A3	Intensity and electric field are related under Poynting’s vector	1.00
A4	Coefficient $\alpha$ of dipole moment $\vec p$ is calculated	1.00
A5	Gradient force averaged over time is derived	1.00
A6	Answer for $\Omega$	0.75
A7	Radiation force $F_{\mathrm{rad}}$ is found and expressed with $\lambda$	1.00
A8	Answer for equilibrium position $x$	0.75
B1	Correct ray diagram and proper deviation angle is derived	2.00
B2	Finds a proper expression for infintessimal force due to momentum change	1.00
B3	Uses law of cosines to obtain a general expression for $\rho$	0.50
B4	Integral expression for force in $x,y,z$ directions is correct	0.75
B5	Integral expressions for torque is correct	0.75
C1	Differential equation is created and solved	1.0
C2	Correct answer is expressed (can be in a piecewise function)	0.50
C3	Accounts for forces due to radiation and is able to show it does not affect rotation	0.50

Acoustic Levitation — (d)

For either case, there are different processes happening which require different analysis.

When $f$ is low

By the condition $f\ll c_s/A$ , the acoustic wave produced by the piston travels much faster than the piston as $c_s\gg fA$ . For a general traveling wave, we define:

\psi(x,t)=A\sin(kx-2\pi ft)

where $\psi$ refers to the longitudinal displacement of the wave and $k$ is the wavenumber. Now consider a part of the gas that is in equilibrium followed by the gas in a later state.

Complete paired before/after gas slices, connecting arrow, solid/dashed boundaries and x, x+Delta x, x+psi(x), x+Delta x+psi(x+Delta x) labels.

We can approximate $\psi(x+\Delta x)$ to be $\psi(x)+\Delta\psi$ . Note that the volume of this slice of gas can be written as $S\Delta x$ but the change in volume of the gas between both successive pictures can be written as $\Delta V=S\Delta\psi$ . Furthermore, the ideal gas equation holds for small amplitudes which means that $pV^\gamma=\mathrm{const}$ , and by logarithmic differentiation,

\frac{\Delta p}{p_0}+\gamma\frac{\Delta V}{V_0}=0.

Thus, we can rewrite the volume terms as:

\frac{\Delta V}{V_0}=\frac{S\Delta\psi}{S\Delta x}\to\frac{\partial\psi}{\partial x}.

The change in pressure of a segment of the gas at $x=0$ can be characterized as

\Delta p(0,t)=-\gamma p_0\left.\frac{\partial\psi}{\partial x}\right|_{x=0}=\gamma p_0Ak\cos(2\pi ft).

Now the force acting on the piston due to this change in pressure can be accurately described as $\Delta F=S\Delta p(0,t)$ . Power can be written as $P=\vec F\cdot\vec v$ and we know that since $\psi=A\cos(2\pi ft)$ , then $v=-2\pi Af\sin(2\pi ft)$ . The wavenumber is written as $k=2\pi/\lambda=2\pi f/c_s$ . The total and average power acting on the piston for each quarter cycle is then,

P=4\pi^2\gamma p_0A^2f^2c_s\cos^2(2\pi ft)\quad\Longrightarrow\quad\langle P\rangle=2\pi^2\gamma f^2p_0A^2c_s.

When $f$ is very high

In the limit of $f\gg c_s/A$ , the piston will start to move much faster than the actual molecules in the air. Therefore, at some point in time, the piston will actually not have ”contact” with the molecular layer. This is because after pushing the gas molecules, there will be an empty void which will eventually be replaced at a distance $d\sim c_s/f$ . The piston will only exert a force (and hence power) upon having contact with the layer of gas. Suppose the piston moves a distance $A-d$ in a time $\tau$ . We can then find that

A\cos(2\pi f\tau)=A-\frac{c_s}f\quad\Longrightarrow\quad2A\sin^2(\pi f\tau)=\frac{c_s}f.

Assuming $\tau$ is small, we can use approximations such that

2\pi^2Af^2\tau^2=\frac{c_s}f\quad\Longrightarrow\quad\tau=\frac1\pi\sqrt{\frac{c_s}{2Af^3}}.

The entire cycle takes a time $t=1/f$ which means we must consider the interval $(\tau,1/f)$ . We can use a similar process as in the analysis for $f\ll c_s/A$ now. Except this time, to find the power used, we will find the energy per cycle and divide by $1/f$ . The potential energy per unit length can be written as

\mathrm dU=\frac12S\gamma p_0\left(\frac{\partial\psi}{\partial x}\right)^2\quad\Longrightarrow\quad\Delta U=\frac12S\gamma p_0\int_\tau^{1/f}\left(\frac{\partial\psi}{\partial t}\right)^2\mathrm dt.

Thus, we must now evaluate:

\Delta U=8S\gamma p_0\pi^4A^2f^4\cos^2\left(2\pi f\left(\frac1f-\tau\right)\right).

Having all constants be defined under $\xi$ , we can now write

\begin{aligned}\Delta U&\approx\xi\left(1-\left(2\pi f\left(\frac1f-\frac1\pi\sqrt{\frac{c_s}{2Af^3}}\right)\right)^2\right)\\&=\xi\left(1-4\pi^2f^2\left(\frac1{f^2}-\frac2\pi\sqrt{\frac{c_s}{2Af^5}}+\frac1{\pi^2}\frac{c_s}{2Af^3}\right)\right).\end{aligned}

The average power used is thus

\langle P\rangle=\frac{\Delta U}{1/f}=8S\gamma p_0\pi^3A^2f^5\left(\frac1{f^2}-\frac2\pi\sqrt{\frac{c_s}{2Af^5}}+\frac1{\pi^2}\frac{c_s}{2Af^3}\right).

(e)

The internal energy of $n$ moles of the gas is given by $U=nC_V\Delta T$ . The power dissipated by the shockwave is then

P=\frac{\mathrm dU}{\mathrm dt}=\frac{\mathrm dn}{\mathrm dt}C_V\Delta T.

Note that $\Delta n=\rho S\Delta x/m$ which means that $\mathrm dn/\mathrm dt=\rho Sc_s/m$ where $c_s$ is the speed of the sound wave. Hence, the power dissipated can be written as $P=3R\rho Sc_s\Delta T/2m$ . Thus, the change in temperature is when both power generation and dissipation is equated. Note that the process is adiabatic as the wavelength of the sound waves are much larger than the mean free path of the gas molecules implying slow motion.

\frac{3R\rho Sc_s\Delta T}{2m}=P_{\mathrm{adb}}\quad\Longrightarrow\quad\Delta T=\frac{2P_{\mathrm{adb}}m}{3R\rho Sc_s}.

(f)

The force on the cylinder will be

F=-\pi R^2(p(y+h)-p(y))=-\pi R^2h\frac{\mathrm dP}{\mathrm dy}.

From part D, we know that

\Delta p(y,t)=\gamma p_0Ak\cos(ky-2\pi ft)\quad\Longrightarrow\quad\frac{\mathrm dp}{\mathrm dy}=\gamma p_0Ak\sin(ky-2\pi ft).

From this, we can see that the cylinder will levitate near the pressure nodes, or where $\mathrm dP/\mathrm dy=0$ . These will typically be compressions as they imply a stable equilibrium. As the waves are travelling, the puck will float up with the sound wave and then fall repeating the same process over again implying oscillations. For true levitate, we would require the waves to be standing.

Grading Scheme

D1	Correct reasoning for what happens to the piston in the limit of $f\ll c_s/A$	0.25
D2	Shows ideal gas equation holds and relates $\Delta p$ and $\Delta V$	1.00
D3	Rewrites $\Delta V/V_0$ and derives power $P(t)$	1.25
D4	Correct expression for average power $\langle P\rangle$	0.75
D5	Correct reasoning for what happens to the piston in the limit of $f\gg c_s/A$	0.50
D6	Finds characteristic time $\tau$ for when piston hits the ”molecular wall”	0.50
D7	Finds potential energy per unit length per cycle	1.25
D8	Correct expression for average power	0.75
E1	Relates internal energy and power dissipation (alternate: uses equipartition theorem)	3.00
E2	Correct answer for $\Delta T$	1.00
F1	Relates force $F$ and pressure gradient $p$ to show sinusoidal nature	1.00
F2	Correct reasoning for how the cylinder levitates and where it does	1.00

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

You are allowed to refer to Wikipedia (but not any other sources). You should typically rederive proofs instead of citing them.
Uploading images: View this link to find how to upload images on overleaf.
You are allowed to mix LaTeX and handwritten PDFs for your answer sheet. Note that you need to write your solution to the problem alongside your final answers. If you desire a LaTeX template, we have made one for you here. Make a copy of the project to edit the document.
You can use basic formulae related to topics on the IPhO syllabus without derivation but all other formulae which go beyond that must be derived.
If you find that space is running out on your LaTeX file, try to use $\verb|\newpage|$ and see if that fixes it.

Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf