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IPhO 2026, Теория — Задача 2. Cooking with sunlight?

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Theoretical Exam -- English (Official) · 10 т.

Условие

The redistribution of luminous flux into regions of high and low intensity due to reflection or refraction can lead to the formation of caustics: curves formed by the intersection of infinitesimally close light rays. As an example, Figure 2a shows the caustic formed by rays incident paraxially on a half-cylindrical mirror. Figure 2b shows a photograph of the caustic formed by the reflection of light on the bottom part of a coffee cup. The brightest and central point of the caustic is called the cusp; at this point, the light intensity is maximal.

Ray diagram: vertical parallel rays (orange) enter a half-circular mirror (blue arc with hatching) and after reflection their envelope forms a caustic with two arcs meeting at a central cusp. — Fig. 2a.

Photograph from above of a white coffee cup with a handle; a bright caustic curve with a central cusp is visible on the bottom of the cup. — Fig. 2b.

Caustics have a variety of applications in science, ranging from high-precision optics and quantum optics to astrophysics. In this question, you will study the caustic formation by light reflection on a half-cylinder mirror of radius $R$ , as shown in figure 2c. You will also investigate other aspects of this mirror, such as its application in solar cookers.

Throughout the problem, assume that the incident rays come from a distant external light source and are parallel to the mirror's optical axis. By symmetry, it is sufficient to study reflection in one plane. Figure 2d shows such a plane, the mirror, and the coordinate system used throughout the problem.

Perspective drawing of a half-cylinder mirror (light blue inner surface) of diameter 2R; orange rays enter parallel to the axis plane and reflect inside it; a dashed line marks the symmetry plane. — Fig. 2c.

T2-A [1.5 pts]. Multiple reflections

Cross-section of the mirror: a half-circle (blue, hatched) from -R to R on the x axis, with y axis vertical; an incident vertical ray (orange arrows) at distance x from the y axis. — Fig. 2d.

Step graph of the number of reflections N (1, 2, 3 on the vertical axis) versus x; N = 1 between -x1 and x1, N = 2 between x1 and x2 (and -x2 and -x1), N = 3 between x2 and x3 (and -x3 and -x2); filled and open circles mark the ends of the intervals. — Fig.2e.

Figure 2e shows the relationship between the $x$ -coordinate of an incident ray and the number $N$ of reflections it undergoes, up to $N = 3$ . As $x$ varies over the interval $(-R, R)$ , the variable $N$ varies over all natural numbers greater than or equal to 1. The maximum distance from the optical axis that allows an incident ray to be reflected by the mirror at most $N$ times is denoted $x_N$ .

T2-A1 Find the general expression for $x_N$ in the infinite sequence $x_1, x_2, x_3, \ldots$ that limits the coordinates of the rays with at most $N$ reflections. Express your answer in terms of $R$ and $N$ . [1,5 т.]

T2-B [4.0 pts]. Solar Cooker

Perspective drawing of the solar cooker: a half-cylinder mirror (light blue) of width 2R on two legs, with a white cylindrical container of radius a whose axis is parallel to the mirror axis, at distance R/2 from the bottom of the mirror. — Fig.2f.

Consider a system consisting of a half-hollow-cylinder mirror and a cylindrical container. The cylinder is fully absorbing: any ray of light hitting it will be absorbed. The mirror radius is $R$ . The container radius $a$ . The axes of the mirror and the container are parallel. The center of the container is $R/2$ from the mirror on the system's symmetry plane, as shown in figure 2f. The mirror harnesses solar radiation to heat food inside the container. Assume that the sunlight is of constant and uniform intensity (power per unit area), and that the light rays are parallel to the optical axis of the mirror. Furthermore, assume that $a$ is such that any ray that is absorbed by the cylinder reflects from the mirror at most once. Let $\theta_{\max}$ be the maximum angle of incidence on the mirror (measured with respect to the normal drawn at the point of incidence) of any reflected ray striking the container.

T2-B1 The container radius is $a = \alpha\sin(\theta_{max}) + \beta\sin(2\theta_{max})$ . Write $\alpha$ and $\beta$ in terms of $R$ . [2 т.]

If you do not find $\alpha$ and $\beta$ , use the expression given for the rest of the problem.

T2-B2 Let $P_0$ be the power that would be received by the cylinder if the mirror was not present. Write $P/P_0$ , the ratio of the actual received power $P$ to $P_0$ , in terms of $\theta_{\max}$ . [1,5 т.]

T2-B3 If $R = 1.0$ m, write the value of $a$ such that $P = 5P_0$ . Give your answer in cm. [0,5 т.]

T2-C [4.5 pts]. Caustics and Cusp

T2-C1 Consider a ray $A$ that strikes the mirror at an angle $\theta$ , as shown in Figure 2g. Upon reflection, the equation corresponding to this ray in the coordinate system defined in Figure 2g. is $y = m_A x + b_A$ . Write $m_A$ and $b_A$ in terms of $\theta$ and $R$

Hint: the expected form of $m_A = K_1\cot(K_2\theta)$ and $b_A = R\frac{K_3}{\cos(K_4\theta)}$ , with $K_i$ numerical constants. [0,5 т.]

Half-circle mirror in the x–y plane with ends at -R and R. A vertical incident ray A at positive x hits the mirror; the dashed radius to the point of incidence makes the angle θ with the vertical ray. The reflected ray goes down-left and is labelled y=m_A x+b_A. — Fig. 2g.

T2-C2 Now consider a ray $B$ , parallel to $A$ , that strikes the surface at an angle $\theta + \Delta\theta$ , with $\Delta\theta \ll \theta$ . Upon reflection, the equation corresponding to this ray is $y = m_B x + b_B$ . Write $m_B$ and $b_B$ in terms of $\theta$ , $\Delta\theta$ , and $R$ , approximating to first order in $\Delta\theta$ . [2 т.]

T2-C3 Find the coordinates of the point of intersection $(X_c, Y_c)$ between the reflected rays of $A$ and $B$ . Write your answer in terms of $R$ and $\theta$ . [1 т.]

T2-C4 For $\theta \ll 1$ , we have $Y_c = v|X_c|^{p/q} + u$ . Write $v$ and $u$ in terms of $R$ and find the values of the integers $p$ and $q$ (equivalently, give the rational number $p/q$ ). [1 т.]

Отговори

Решение

PART A [1.5 pts]. Multiple reflections

T2-A1 [1.5 pts]. For $n = 1$ , finding $x_1$ corresponds to considering the limiting case of the ray, shown in blue in the figure below.

Half-circle mirror with center O on the x axis; a vertical blue incident ray from A hits the mirror at C1, and the red reflected ray passes through the edge point B; the dashed radius OC1 makes angle θ1 with the vertical; angles θ1 are marked at C1, at O and at B.

Limiting case for two reflections: vertical blue ray from A2 hits the mirror at C2, reflects to D2 and then to the edge point B; dashed radii OC2 and OD2; the base angles θ2 are marked at C2, D2, O and B.

In this case, the incident ray passes through point $A$ , and the reflected ray, shown in red in the figure, passes exactly through point $B$ on the edge of the mirror.

Let $\theta_1$ be the angle formed by the radius passing through the reflection point $C_1$ with the vertical axis (the $y$ axis). Let $O$ be the center of the mirror.

Since the incident ray is parallel to the $y$ axis, the angle of incidence is

\angle OC_1A = \theta_1\,.

By the law of reflection,

\angle OC_1A = \angle OC_1B = \theta_1\,.

Additionally, note that the triangle $C_1BO$ is isosceles, so

\angle C_1BO = \theta_1\,.

Finally, summing the internal angles of the triangle $C_1BA$ , we have

3\theta_1 + \frac{\pi}{2} = \pi,

\theta_1 = \frac{\pi}{6}.

Correspondingly, from the triangle $OC_1A$ , we obtain

\boxed{x_1 = R\sin\frac{\pi}{6} = \frac{R}{2}.}

For $n = 2$ , the limiting case for two reflections is shown in the figure below.

In analogy with the case $n = 1$ , we let $\theta_2$ be the angle between the radius passing through the reflection point $C_2$ and the mirror axis. We see that the trajectory of the reflected ray determines two isosceles triangles with a common side and a base angle of $\theta_2$ . Also note that the ray's trajectory forms half of a regular pentagon. Summing over the internal angles of the two isosceles triangles and the rectangular triangle, we obtain

5\theta_2 + \frac{\pi}{2} + \pi = 3\pi,

which simplifies to

\theta_2 = \frac{3\pi}{10}.

Correspondingly,

\boxed{x_2 = R\sin\frac{3\pi}{10} = \frac{1+\sqrt{5}}{4}R \approx 0.8R.}

Case n=3: For completeness, we present the limiting case for 3 reflections in the figure below.

Limiting case for three reflections: vertical blue ray from A3 hits the mirror at C3, reflects to D3, then to E3 and to the edge point B; dashed radii from O; angles θ3 marked at each reflection point and at B.

Letting $\theta_3$ be the incidence angle, we observe that the trajectory of the reflected ray determines three isosceles triangles with common sides and a base angle of $\theta_3$ . Also notice that the ray's trajectory now forms half of a regular heptagon. Summing over the internal angles of the three isosceles triangles and the rectangular triangle, we obtain

Limiting case for the solar cooker: grey circle (container) with center Oc on the y axis; a vertical blue ray from A reflects at B, and the red reflected ray is tangent to the container at C and crosses the y axis at D; angles θmax marked at B and at O.

7\theta_3 + \frac{\pi}{2} + \pi = 4\pi,

which simplifies to

\theta_3 = \frac{5\pi}{14}

Correspondingly

\boxed{x_3 = R\sin\frac{5\pi}{14} \approx 0.9R.}

Generalizing the result, the limiting case for $n$ reflections corresponds to the ray with incidence angle $\theta_N$ .

The trajectory of the reflected ray determines $n$ isosceles triangles with common sides and a base angle of $\theta_N$ .

Also note that the ray's trajectory now forms half of a regular $2n+1$ -gon. Summing over the internal angles of the $n$ isosceles triangles and the rectangular triangle determined by the incident ray, we obtain

(2n\theta_N + \pi) + \left(\frac{\pi}{2} + \theta_N\right) = (n+1)\pi,

(2n+1)\theta_N + \frac{3\pi}{2} = (n+1)\pi,

which simplifies to

\theta_N = \frac{2n-1}{4n+2}\pi

Correspondingly

\boxed{x_n = R\sin\frac{2N-1}{4N+2}\pi = R\cos\frac{\pi}{2N+1}}

T2-B [4.0 pts]. Solar Cooker

T2-B1 [2.0 pts].

The limiting case is shown in the figure below.

$O_c$ is the center of the container.
$A$ is the incidence point on the $x$ -axis
$B$ is the reflection point
The reflected ray is tangential to the container at the point $C$
The reflected ray crosses the $y$ -axis at $D$ .

As the triangle $ODB$ is isosceles,

\angle ODB = \pi - 2\theta_{\max}\,,

\cos\theta_{\max} = \frac{R}{2\overline{OD}}.

\overline{OD} = \frac{R}{2\cos\theta_{\max}}.

The triangle $O_cDC$ is rectangle

\begin{aligned} a &= \overline{O_cD}\sin\left(\pi - 2\theta_{\max}\right) \\ &= \left(\overline{OD} - \frac{R}{2}\right)\sin\left(2\theta_{\max}\right) \\ &= \left(\frac{R}{2\cos\theta_{\max}} - \frac{R}{2}\right)\sin\left(2\theta_{\max}\right) \\ &= R\sin\theta_{\max} - \frac{R}{2}\sin\left(2\theta_{\max}\right) \end{aligned}

The coefficients read

\boxed{\alpha = R}

\boxed{\beta = -\frac{R}{2}}

T2-B2 [1.5 pts]. Without the mirror:

P_0 = 2Ial,

where $I$ is the sunlight intensity and $l$ is the length of the cylinder.
With the mirror:

\begin{aligned} P &= 2I\overline{OA}l \\ &= 2IRl\sin\theta_{\max} \end{aligned}

Thus

\begin{aligned} \frac{P}{P_0} &= \frac{2IRl\sin\theta_{\max}}{2Ial} \\ &= \frac{R\sin\theta_{\max}}{R\sin\theta_{\max}(1-\cos\theta_{\max})} \end{aligned}

\boxed{\frac{P}{P_0} = \frac{1}{1-\cos\theta_{\max}}.}

Alternative answer in terms of $\alpha$ and $\beta$ : Without the mirror:

P_0 = 2Ial = 2I(\alpha\sin\theta_{\max} + \beta\sin\left(2\theta_{\max}\right))l

where $I$ is the sunlight intensity and $l$ is the length of the cylinder.

With the mirror:

\begin{aligned} P &= 2I\overline{OA}l \\ &= 2IRl\sin\theta_{\max} \end{aligned}

Thus

\begin{aligned} \frac{P}{P_0} &= \frac{2IRl\sin\theta_{\max}}{2Ial} \\ &= \frac{R\sin\theta_{\max}}{\alpha\sin\theta_{\max} + \beta\sin\left(2\theta_{\max}\right)} \end{aligned}

\boxed{\frac{P}{P_0} = \frac{R}{\alpha + 2\beta\cos\theta_{\max}}.}

T2-B3 [0.5 pts]. If $P = 5P_0$

5 = \frac{1}{1-\cos\theta_{\max}},

\cos\theta_{\max} = \frac{4}{5},

\sin\theta_{\max} = \frac{3}{5}.

Finally

a = (1\mathrm{m})\frac{3}{5}\left(1 - \frac{4}{5}\right).

\boxed{a = 12\,\mathrm{cm}.}

Alternative solution parametrized in terms of $\alpha$ and $\beta$

a = -\frac{R\sqrt{100\beta^2 - (R-5\alpha)^2}}{50\beta}

T2-C [4.5 pts]. Caustics and Cusp

T2-C1 [0.5 pts].

Half-circle mirror with vertical blue ray from A reflected at B; the red reflected ray crosses the y axis at C; dashed radius OB; angles θ marked at B and at O.

The reflected ray forms an angle $2\theta$ with the vertical axis. Therefore, the slope of the line is

\boxed{m_A = \cot(2\theta).}

The reflected ray intersects the $y$ -axis at $C$ , $\overline{OC} = b_A$ , forming the isosceles triangle $OBC$ . from the intersection

\cos\theta = \frac{R}{2b_A}.

\boxed{b_A = \frac{R}{2\cos\theta}.}

Equivalent expressions:

\begin{aligned} m_A &= \cot(2\theta) = \frac{1}{\tan(2\theta)} = \frac{\cot(\theta)}{2} - \frac{\tan(\theta)}{2} \\ b_A &= \frac{R}{2\cos\theta} = \frac{R}{2}\sec\theta \end{aligned}

T2-C2 [2.0 pts].

The slope is

\begin{aligned} m_B &= \cot(2\theta + 2\Delta\theta) \\ &= \frac{\cot(2\theta)\cot(2\Delta\theta) - 1}{\cot(2\theta) + \cot(2\Delta\theta)} \\ &= \frac{\cot(2\theta) - \tan(2\Delta\theta)}{\cot(2\theta)\tan(2\Delta\theta) + 1} \\ &= (\cot(2\theta) - 2\Delta\theta)(1 - 2\Delta\theta\cot(2\theta)) + \ldots \\ &= \cot(2\theta) - 2\Delta\theta(1 - \tan^2(2\Delta\theta)) + \ldots \end{aligned}

Then, the first-order expansion of $m_B$ reads

\boxed{m_B = \cot(2\theta) - 2\csc^2(2\theta)\Delta\theta.}

On the other hand, the intercept is

\begin{aligned} b_B &= \frac{R}{2}\sec(\theta + \Delta\theta) \\ &= \frac{R}{2}\frac{\sec(\theta)\sec(\Delta\theta)}{1 - \tan(\theta)\tan(\Delta\theta)} \\ &= \frac{R}{2}\left(\sec(\theta)(1 + \tan(\theta)\Delta\theta)\right) + \ldots \end{aligned}

and the first-order expansion reads

\boxed{b_B = \frac{R}{2}\sec(\theta)\left(1 + \tan(\theta)\Delta\theta\right)}

Equivalent expressions:

\begin{aligned} m_B &= \cot(2\theta) - 2\csc^2(2\theta)\Delta\theta \\ &= \frac{1}{2}\csc^2(2\theta)(\sin(4\theta) - 4\Delta\theta) \\ &= \frac{1}{2}\left(\cot(\theta) - \tan(\theta) - \csc^2(\theta)\sec^2(\theta)\right)\Delta\theta \end{aligned}

\begin{aligned} b_B &= \frac{R}{2}\sec(\theta)\left(1 + \tan(\theta)\Delta\theta\right) \\ &= \frac{R}{2\cos(\theta)}\left(1 + \tan(\theta)\Delta\theta\right) \end{aligned}

T2-C3 [1.0 pts].

The coordinates of the intersection point, $(X_C, Y_C)$ , fulfill the following equations:

\begin{aligned} Y_C &= m_A X_C + b_A \\ Y_C &= m_B X_C + b_B, \end{aligned}

Solving:

(X_C, Y_C) = \left(-\frac{b_A - b_B}{m_A - m_B}, \frac{m_A b_B - b_A m_B}{m_A - m_B}\right)

Replacing $m_B$ and $b_B$ , we find:

(X_C, Y_C) = \left(R\sin^3(\theta), \frac{R}{2}\cos(\theta)(2 - \cos(2\theta))\right)

\boxed{X_C = R\sin^3(\theta)}

\boxed{Y_C = \frac{R}{2}\cos(\theta)(2 - \cos(2\theta))}

Equivalent expressions:

\begin{aligned} Y_C &= \frac{R}{2}\cos(\theta)(2 - \cos(2\theta)) \\ &= \frac{R}{4}(3\cos(\theta) - \cos(3\theta)) \\ &= \frac{R}{4}(3\cos(\theta) - \cos^3(\theta) + 3\sin^2(\theta)\cos(\theta)) \end{aligned}

T2-C4 [1.0 pts]. For $\theta \ll 1$ , we use the approximations

\sin\theta \approx \theta,

\cos\theta \approx 1 - \frac{\theta^2}{2}.

Thus, near the cusp, the caustic is approximated as

(X_C, Y_C) \approx \left(R\theta^3, \frac{R}{4}(2 + 3\theta^2)\right)

(X_C, Y_C) \approx \left(R\theta^3, \frac{R}{2} + \frac{3R}{4}\theta^2\right)

Solving for $\theta$ , we get

\theta = \left(\frac{X_C}{R}\right)^{\frac{1}{3}},

and substitute in $Y_C$

Y_C = \frac{R}{2} + \frac{3R}{4}\left(\frac{X_C}{R}\right)^{\frac{2}{3}}

Comparing with

Y_C = v|X_C|^{p/q} + u,

we find

\boxed{v = \frac{3}{4}R^{\frac{1}{3}}}

\boxed{u = \frac{R}{2},}

\boxed{p = 2}

\boxed{q = 3}

This shows that the shape of the cusp is universal and that its tip is located at $y = \frac{R}{2}$ .

Half-circle mirror with the caustic drawn in red (two arcs meeting at a cusp on the y axis) and two nearby vertical rays with their reflected rays intersecting close to the caustic.

Оригинал в Архива: T2.pdf · официални решения: theory_sol.pdf