IPhO 2026 · Theory problem 2

Cooking with sunlight?

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QUESTION T2 [10 pts]. Cooking with sunlight?

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.

Figure 1

Fig. 2a.

Figure 2

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.

Figure 3

Fig. 2c.

Figure 4

Fig. 2d.

T2-A [1.5 pts]. Multiple reflections

Figure 5

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 [1.5 pts]. 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 \(N\).

T2-B [4.0 pts]. Solar Cooker

Figure 6

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 [2.0 pts]. The container radius is \(a=\alpha\sin(\theta_{\max})+\beta\sin(2\theta_{\max})\). Write \(\alpha\) and \(\beta\) in terms of \(R\).

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

T2-B2 [1.5 pts]. Let \(P_0\) be the power that would be received by the cyllinder if the mirror was not present. Write \(P/P_0\), the ratio of the actual received power \(P\) to \(P_0\), in terms of \(\theta_{\max}\).

T2-B3 [0.5 pts]. If \(R=1.0\ \mathrm{m}\), write the value of \(a\) such that \(P=5P_0\). Give your answer in cm.

T2-C [4.5 pts]. Caustics and Cusp

T2-C1 [0.5 pts]. 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 \(y=m_Ax+b_A\). 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.

Figure 7

Fig. 2g.

T2-C2 [2.0 pts]. 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 \(y=m_Bx+b_B\). Write \(m_B\) and \(b_B\) in terms of \(\theta\), \(\Delta\theta\), and \(R\), approximating to first order in \(\Delta\theta\).

T2-C3 [1.0 pts]. 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 \(\theta\).

T2-C4 [1.0 pts]. For \(\theta\ll1\), 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 \(p/q\)).

Official English text of the problem.

Answers

  • T2-A1: \[ \boxed{x_N=R\cos\!\left(\frac{\pi}{2N+1}\right)},\qquad N=1,2,3,\ldots. \]

  • T2-B1: \[ \boxed{\alpha=R,\qquad \beta=-\frac R2}. \]

  • T2-B2: \[ \boxed{\frac{P}{P_0}=\frac{1}{1-\cos\theta_{\max}}}. \]

  • T2-B3: \[ \boxed{a=12\ \mathrm{cm}}. \]

  • T2-C1: \[ \boxed{m_A=\cot(2\theta),\qquad b_A=\frac{R}{2\cos\theta}}. \]

  • T2-C2: To first order in \(\Delta\theta\), \[ \boxed{m_B\simeq\cot(2\theta)-\frac{2\Delta\theta}{\sin^2(2\theta)}}, \qquad \boxed{b_B\simeq\frac{R}{2\cos\theta} +\frac{R\sin\theta}{2\cos^2\theta}\Delta\theta}. \]

  • T2-C3: In the neighboring-ray limit \(\Delta\theta\to0\), \[ \boxed{X_c=R\sin^3\theta,\qquad Y_c=\frac R2\cos\theta\left(1+2\sin^2\theta\right)}. \]

  • T2-C4: \[ \boxed{v=\frac34R^{1/3},\qquad u=\frac R2,\qquad p=2,\quad q=3}. \]

Solution

T2-A1

By symmetry, consider an incident ray with \(x\ge0\). Let \(\phi\) be the polar angle of its first point of incidence, measured from the positive \(x\)-axis. Then \[ x=R\cos\phi,\qquad 0<\phi\le\frac{\pi}{2}. \] Since the incident ray travels vertically, its angle of incidence relative to the normal is \[ \theta=\frac{\pi}{2}-\phi. \]

The law of reflection makes the reflected ray form the same angle \(\theta\) with the radial normal. A chord connecting two successive points of reflection therefore subtends the central angle \[ \pi-2\theta=2\phi. \] The triangle formed by that chord and the two radii is isosceles, so the angle of incidence remains the same at successive reflections.

Consequently, on the corresponding full circle, successive contact points have polar angles \[ \phi,\quad 3\phi,\quad 5\phi,\quad\ldots,\quad (2j-1)\phi. \] Only the upper semicircle is a mirror. The ray exits through the diameter when the next full-circle contact would be on or below it.

At the limiting incident coordinate \(x_N\), the ray undergoes \(N\) reflections and then reaches the far rim. Thus the would-be \((N+1)\)-st contact is at polar angle \(\pi\): \[ (2N+1)\phi_N=\pi. \] It follows that \[ \boxed{x_N=R\cos\!\left(\frac{\pi}{2N+1}\right)}. \]

At equality, the ray exits at the rim, consistent with the endpoints shown in Fig.2e. The first values are \[ x_1=\frac R2,\qquad x_2\approx0.8090R,\qquad x_3\approx0.9010R, \] and \(x_N\to R\) as \(N\to\infty\).

T2-B1

Use the coordinates of Fig.2d. The container centre is \[ C=\left(0,\frac R2\right). \] A ray with incidence angle \(\theta\ge0\) strikes the mirror at \[ M=(R\sin\theta,R\cos\theta). \] The outward unit normal there is \[ \hat{\mathbf n}=(\sin\theta,\cos\theta), \] and the incident propagation direction is \(\hat{\mathbf k}=(0,1)\).

The vector form of the law of reflection gives the reflected unit direction: \[ \begin{aligned} \hat{\mathbf u} &=\hat{\mathbf k} -2(\hat{\mathbf k}\cdot\hat{\mathbf n})\hat{\mathbf n}\\ &=(0,1)-2\cos\theta(\sin\theta,\cos\theta)\\ &=(-\sin2\theta,-\cos2\theta). \end{aligned} \]

The perpendicular distance from \(C\) to this reflected line is the magnitude of the two-dimensional cross product of \(C-M\) with \(\hat{\mathbf u}\): \[ \begin{aligned} d(\theta) &=\left|(C_x-M_x)\hat u_y-(C_y-M_y)\hat u_x\right|\\ &=\left|-R\sin\theta+\frac R2\sin2\theta\right|\\ &=R\sin\theta(1-\cos\theta). \end{aligned} \] For \(0\le\theta<\pi/2\), this distance increases with \(\theta\), since \[ d'(\theta)=R(1-\cos\theta)(1+2\cos\theta)\ge0. \]

The limiting accepted reflected ray is tangent to the container, so \(d(\theta_{\max})=a\). Therefore \[ a=R\sin\theta_{\max}-\frac R2\sin2\theta_{\max}. \] Comparison with the given expression yields \[ \boxed{\alpha=R,\qquad\beta=-\frac R2}. \]

T2-B2

Let \(I\) be the uniform incident intensity and \(L\) a common illuminated length along the cylinder axes. Without the mirror, the container presents projected area \(2aL\), so \[ P_0=2aLI. \]

The greatest incoming distance from the optical axis that leads to absorption is \[ h_{\max}=R\sin\theta_{\max}. \] Rays with \(|x|\le a\) strike the container directly. Rays with \[ a<|x|\le h_{\max} \] reach the mirror and are then absorbed after one reflection.

The directly absorbed rays shadow the corresponding portion of the mirror. Thus the direct and reflected contributions are \[ P_{\mathrm{direct}}=2aLI, \qquad P_{\mathrm{reflected}}=2(h_{\max}-a)LI. \] By the stated restriction, no additional absorbed power comes from rays undergoing two or more reflections. For the ideal mirror, conservation of power therefore gives \[ P=2h_{\max}LI=2R\sin\theta_{\max}LI. \] Hence \[ \frac{P}{P_0} =\frac{R\sin\theta_{\max}}{a}. \] Substituting the result of T2-B1, \[ \boxed{\frac{P}{P_0}=\frac{1}{1-\cos\theta_{\max}}}. \]

T2-B3

The condition \(P=5P_0\) gives \[ 1-\cos\theta_{\max}=\frac15, \qquad \cos\theta_{\max}=\frac45. \] Thus \[ \sin\theta_{\max}=\frac35 \] and \[ a=R\sin\theta_{\max}(1-\cos\theta_{\max}) =R\frac35\frac15 =\frac{3R}{25}. \] With \(R=1.0\ \mathrm m\), \[ \boxed{a=0.12\ \mathrm m=12\ \mathrm{cm}}. \]

T2-C1

From T2-B1, the reflected ray passes through \[ M=(R\sin\theta,R\cos\theta) \] and has direction \[ \hat{\mathbf u}=(-\sin2\theta,-\cos2\theta). \] Its slope is therefore \[ \boxed{m_A=\frac{\hat u_y}{\hat u_x}=\cot2\theta}. \]

Using the coordinates of \(M\), \[ \begin{aligned} b_A &=R\cos\theta-R\sin\theta\cot2\theta\\ &=R\cos\theta-\frac{R\cos2\theta}{2\cos\theta}\\ &=\boxed{\frac{R}{2\cos\theta}}. \end{aligned} \] In the notation of the hint, \[ K_1=1,\qquad K_2=2,\qquad K_3=\frac12,\qquad K_4=1. \]

The central reflected ray, at \(\theta=0\), is the vertical line \(x=0\). Results involving its slope are understood through the limit \(\theta\to0\).

T2-C2

For ray \(B\), replace \(\theta\) by \(\theta+\Delta\theta\). Taylor expansion to first order gives \[ m_B=m_A+\frac{dm_A}{d\theta}\Delta\theta +O\!\left((\Delta\theta)^2\right). \] Since \[ \frac{d}{d\theta}\cot2\theta=-2\csc^2 2\theta, \] we obtain \[ \boxed{ m_B\simeq\cot2\theta-\frac{2\Delta\theta}{\sin^2 2\theta} }. \]

Similarly, \[ \frac{d}{d\theta}\left(\frac{R}{2\cos\theta}\right) =\frac{R\sin\theta}{2\cos^2\theta}, \] so \[ \boxed{ b_B\simeq\frac{R}{2\cos\theta} +\frac{R\sin\theta}{2\cos^2\theta}\Delta\theta }. \] These expansions retain only terms through first order in \(\Delta\theta\), as requested.

T2-C3

At the intersection of the two reflected rays, \[ m_AX_c+b_A=m_BX_c+b_B. \] Consequently, \[ (m_B-m_A)X_c+(b_B-b_A)=0. \] Using the first-order differences from T2-C2 and taking the neighboring-ray limit, \[ -\frac{2X_c}{\sin^2 2\theta} +\frac{R\sin\theta}{2\cos^2\theta}=0. \] Therefore \[ X_c=\frac{R\sin\theta\,\sin^2 2\theta}{4\cos^2\theta}. \] Since \(\sin2\theta=2\sin\theta\cos\theta\), \[ \boxed{X_c=R\sin^3\theta}. \]

Substitution into ray \(A\)'s equation gives \[ \begin{aligned} Y_c &=R\sin^3\theta\cot2\theta+\frac{R}{2\cos\theta}\\ &=\frac{R}{2\cos\theta} \left(1+\sin^2\theta\cos2\theta\right)\\ &=\boxed{\frac R2\cos\theta\left(1+2\sin^2\theta\right)}. \end{aligned} \] Here the simplification uses \[ 1+\sin^2\theta\cos2\theta =\cos^2\theta\left(1+2\sin^2\theta\right). \]

These parametric equations describe the caustic. The branch on the other side of the optical axis follows by symmetry: \(X_c\) changes sign while \(Y_c\) remains unchanged.

T2-C4

Near the optical axis, use the small-angle expansions \[ \sin\theta=\theta+O(\theta^3), \qquad \cos\theta=1-\frac{\theta^2}{2}+O(\theta^4). \] The caustic coordinates then become \[ X_c=R\theta^3+O(R\theta^5) \] and \[ \begin{aligned} Y_c &=\frac R2 \left(1-\frac{\theta^2}{2}+O(\theta^4)\right) \left(1+2\theta^2+O(\theta^4)\right)\\ &=\frac R2+\frac{3R}{4}\theta^2+O(R\theta^4). \end{aligned} \] For both symmetric branches, \[ \theta^2\simeq\left(\frac{|X_c|}{R}\right)^{2/3}. \] Eliminating \(\theta\) therefore gives, to leading order near the cusp, \[ \boxed{ Y_c\simeq\frac R2+\frac34R^{1/3}|X_c|^{2/3} }. \] Thus \[ \boxed{ v=\frac34R^{1/3},\qquad u=\frac R2,\qquad p=2,\qquad q=3 }. \]

The cusp is at \((0,R/2)\). Its two symmetric branches open toward increasing \(y\) and share a vertical tangent there.

The solution was checked against the official solution.