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EuPhO 2025, Теория — Задача 1. Sunny

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Theoretical Problems. Language: English · 10 т.

Условие

You are asked to study the features of the brightly lit circle and dark rings in the figures below. Make your calculations for an idealized situation: the chair leg is strictly cylindrical of radius $a$ , strictly vertical, with a perfectly smooth, cylindrical, and perfectly reflecting surface. You may make any additional model assumptions and approximations you deem reasonable that will simplify your calculations.

a) Determine how the illuminance surplus $I(r,\theta)$ inside the brightly lit circle on the floor depends on the polar coordinates $r \gg a$ and $\theta$ . The illuminance quantifies the amount of incoming light per area. By “surplus” we mean the additional illuminance introduced due to the presence of the cylinder. Express the answer in terms of $I_0$ defined as the illuminance difference between points A and B in the figure. [5 т.]

Photo of a bright circular pattern with dark rings on a wooden floor around a vertical cylindrical chair leg; labels: point B, point A, radius r from the leg, I(r,θ), and angle θ.

b) In the following figure some fingers are blocking some of the light from reaching the chair leg. Let $R(\theta)$ denote the radial distance of the middle dark ring as a function of the angle $\theta$ and let $R_{\min}$ be the minimal value of $R(\theta)$ . Determine $R(\theta) - R_{\min}$ . [5 т.]

Photo of the bright circle and dark rings on a wooden floor, with a hand whose fingers reach down from the top and block part of the light; labels R(θ) and angle θ near the middle dark ring.

Решение

Top view: a circle of radius a (the chair leg) touched by an incoming horizontal ray; angles α, β and θ are marked between the ray, the dashed radius a and the reflected ray l; the reflected ray ends in a red dot labelled 'Illuminated spot on floor'; distance x marked at left, e at the bottom. — Figure 1: Top view of a light ray hitting the chair leg. The incoming light ray strikes the cylinder at the blue dot, and the reflected ray strikes the floor at the red dot.

Side view: a rectangle of width a·cosα and height h (with h0 marked), Centerline dashed vertically; a ray with angle γ enters at a blue dot, reflects and leaves to a red dot at lower right with horizontal offset l. — Figure 2: Side view, with the reflected light ray rotated to lie in the same plane as the incident ray.

Part a) Since points A and B are very close, $I_0$ is the illuminance surplus due to direct sunlight.

Consider a small (infinitesimal) horizontal rectangle $[x, x+\delta x] \times [y, y+\delta y]$ placed in the light beam. As depicted in Fig 1, we let $l$ be the horizontal distance between the point where a ray hits the cylinder and where it hits the floor and $\alpha$ the horizontal angle between the ray and the surface normal. The spot on the floor made by the light makes an angle of $2\alpha$ from the incident ray, and forms a small patch of size $(l\,2\delta\alpha)(\delta l)$ on the floor.

Considering the side view sketched in Figure 2, we see that $\delta l \propto \delta y$ for both the ray that hits the ground directly, or a ray that is reflected before hitting the ground.

We thus have

I_0\delta x = 2Il\delta\alpha \qquad (1)

Let $\beta = \pi - \alpha$ as sketched in Fig 1. Then

\delta x = a\,(\sin(\beta+\delta\beta) - \sin\beta) \approx a\cos\beta\,\delta\beta \qquad (2)

Therefore,

I = -\frac{I_0 a}{2l}\cos(\beta) \qquad (3)

(Note that $\cos\beta < 0$ .) This expression is accurate to order $(a/l)^2$ , but is not in terms of the variables requested. A more accurate expression would be

I = -\frac{I_0 a}{\sqrt{4l^2 + a^2\sin^2(\beta)}}\cos(\beta),

but it is not necessary to find this expression.

To obtain $I$ as a function of $r$ and $\theta$ , we need to express $\beta$ and $l$ in terms of $r$ and $\theta$ .

If $a \ll r$ , $l$ and $r$ almost coincide. To leading order thus $l \approx r$ and $\theta \approx \pi - 2\alpha = 2\beta - \pi$ . Substituting $l$ and $\beta$ in (3) results in

I \approx \frac{I_0 a}{2r}\sin(\theta/2) \qquad (4)

Part b) The rings appear because the fingers block the light. For the middle ring, the finger is approximately horizontal.

For $l_0$ be the horizontal distance $l$ for $\alpha = \pi/2$ . Consider the sketch in Fig. 2. Note that

l = l_0 + a\cos\alpha = l_0 + a|\cos(\beta)| \qquad (5)

The minimum value $R_{\min}$ is attained at $\alpha = \beta = \pi/2$ , where $\theta \approx 0$ and $R_{\min} \approx l_0$ within an error of order $(a/r)^2$ . Hence,

l \approx R_{\min} + a\cos\alpha = R_{\min} + a|\cos(\beta)|

The Cosine theorem applied to the triangle between the center of the chair leg, the point where the ray reflects form the leg and the point where it hits the floor, gives the relation

r^2 = a^2 + l^2 - 2al\cos(\beta) \qquad (6)

Using (6) and expanding, we obtain

r^2 = R_{\min}^2 + 4aR_{\min}\cos\alpha + 3a^2\cos^2\alpha

Using $\cos\alpha \approx \sin(\theta/2)$ , and dropping terms of order $(a/r)^2$ , we arrive at

R - R_{\min} \approx 2a\sin(\theta/2) \qquad (7)

The factor of 2 is significant!

Marking scheme T1

Problem T1 a)

		Pts
A	Correct angles, $2\alpha + \theta > \pi$ — Only state $2\alpha + \theta = \pi$ without justification	0.5 / 0.2
B	Reflection law (stated in any way)	0.5
C	$x = a\sin\alpha$	0.5
D	Eq. 1 — Missing factor of 2: -0.3; $r$ instead of $l$ : -0.3	1.0
E	Eq. 3 for $I$	0.5
F	Justify $l \approx r$ — Assume $l \approx r$ without justification	0.5 / 0.2
G	Eq. 4 for $I$	1.5
	Total on T1 a)	5.0

Problem T1 b)

		Pts
H	Cosine Law expression	1.0
I	Eq. 5 for $l$ — qualitative understanding why $R(\theta)$ varies with $\theta$	1.0 / 0.5
J	Justifying $l_0 \approx R_{\min}$ to leading order — Only stating $l_0 \approx R_{\min}$	1 / 0.5
K	Final Eq. 7 — Writing $R - R_{\min} = a\sin(\theta/2)$ : 1.0; Only has $R_{\max} - R_{\min} = 2a$ : 0.5; Only has $R_{\max} - R_{\min} = a$ : 0.0	2.0
	Total on T1 b)	5.0

General rules for marking in T1:

The grain size for marking is 0.1 Pts.
Yellow shaded categories receive a single mark
Partial marks can be awarded for most aspects.
For each mistake in calculation (algebraic or numeric) 0.3 Pts. are deducted.
If a mistake leads to a dimensionally incorrect expression no marks are given for the result.
Propagating errors are not punished again unless they are dimensionally wrong or entail oversimplified/wrong physics (e.g. neglecting friction effects).
Getting a correct answer for a later part that could only be arrived at by doing a previous part correctly will result in the minimum score(s) for previous part(s) that would support the end result.
Justifying F can be done graphically, algebraically, with words, but needs more than just writing $r \approx l$ .
Arriving at a correct G without supporting work would score 4.1 for A-G as (0.2, 0.5, 0.5, 0.7, 0.5, 0.2, 1.5).
Arriving at an incorrect G without supporting work would score 0 for A-G.
Arriving at a correct K without supporting work would score 4.5 for L-O as (1.0, 1.0, 0.5, 2.0)
Arriving at an incorrect K without supporting work would score 0 for H-J, and then the points for K.

Оригинал в Архива: eupho2025_theory_problems.pdf · официални решения: eupho2025_theory_solutions.pdf