Съдържание

IPhO 1971, Теория и експеримент — Задача 4. Question 4.

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 1971

V International Physics Olympiad, 1971. Sofia, Bulgaria

Внимание: Бележка към темата

Решенията са архивна адаптация на Victor Ivanov, както е посочено в оригинала. Формули (1.13), (1.17) и (4.6), както и повтореното означение P₂ в числените резултати на задача 2, са запазени според източника. Добавеният множител g в краткия отговор на задача 1 е отделно обозначена редакционна поправка; текстът на архивното решение остава непроменен.

Source

The problems and the solutions are adapted by Victor Ivanov University of Sofia, Faculty of Physics, 5 James Bourchier Blvd., 1164 Sofia, Bulgaria

Reference: O. F. Kabardin, V. A. Orlov, in “International Physics Olympiads for High School Students”, eds. V. G. Razumovski, Moscow, Nauka, 1985. (In Russian).

Theoretical problems

Solutions — source

Solutions to the problems of the 5-th International Physics Olympiad, 1971, Sofia, Bulgaria

The problems and the solutions are adapted by Victor Ivanov Sofia State University, Faculty of Physics, 5 James Bourcier Blvd., 1164 Sofia, Bulgaria

Reference: O. F. Kabardin, V. A. Orlov, in “International Physics Olympiads for High School Students”, eds. V. G. Razumovski, Moscow, Nauka, 1985. (In Russian).

Theoretical problems

Условие

A spherical aquarium, filled with water, is placed in front of a flat vertical mirror. The radius of the aquarium is $R$ , and the distance between its center and the mirror is $3R$ . A small fish, which is initially at the point nearest to the mirror, starts to move with velocity $v$ along the wall. An observer looks at the fish from a very large distance along a horizontal line passing trough the center of the aquarium.

What is the relative velocity $v_{\mathrm{rel}}$ at which the two images of the fish seen by the observer will move apart? Express your answer in terms of $v$ . Assume that:

The wall of the aquarium is made of a very thin glass.
The index of refraction of water is $n = 4/3$ .

Отговори

Решение

In a small time interval $\Delta t$ the fish moves upward, from point $A$ to point $B$ , at a small distance $d = v\Delta t$ . Since the glass wall is very thin we can assume that the rays leaving the aquarium refract as if there was water – air interface. The divergent rays undergoing one single refraction, as show in Fig. 4.1, form the first, virtual, image of the fish. The corresponding vertical displacement $A_1B_1$ of that image is equal to the distance $d_1$ between the optical axis $a$ and the ray $b_1$ , which leaves the aquarium parallel to $a$ . Since distances $d$ and $d_1$ are small compared to $R$ we can use the small-angle approximation: $\sin\alpha \approx \tan\alpha \approx \alpha$ (rad). Thus we obtain:

(4.1) $d_1 \approx R\,\alpha$ ;

(4.2) $d \approx R\,\gamma$ ;

(4.3) $\alpha + \gamma = 2\beta$ ;

(4.4) $\alpha \approx n\beta$ .

From equations (4.1) - (4.4) we find the vertical displacement of the first image in terms of $d$ :

(4.5)

d_1 = \frac{n}{2 - n}d,

and respectively its velocity $v_1$ in terms of $v$ :

(4.6)

v_1 = \frac{n}{2 - n} = 2v.

The spherical aquarium with optical axis a (dash-dot line); the fish moves from A to B by d; a ray from B refracts at the wall and leaves parallel to the axis as ray b1 at distance d1; the virtual image A1B1 lies to the right; angles α, β, γ are marked. — Fig. 4.1

The rays, which are first reflected by the mirror, and then are refracted twice at the walls of the aquarium form the second, real image (see Fig. 4.2). It can be considered as originating from the mirror image of the fish, which move along the line $A'B'$ at exactly the same distance $d$ as the fish do.

The aquarium in front of a flat mirror: the fish AB of height d and its mirror image A′B′ at distance 4R; rays from the mirror image pass through the aquarium, refract twice and form the real image A2B2 with displacement d2 on the far left, the ray b2 parallel to the axis a; angles α, β, γ and δ are marked. — Fig. 4.2

The vertical displacement $A_2B_2$ of the second image is equal to the distance $d_2$ between the optical axis $a$ and the ray $b_2$ , which is parallel to $a$ . Again, using the small-angle approximation we have:

(4.7) $d' \approx 4R\delta - d$ ,

(4.8) $d_2 \approx R\alpha$

Following the derivation of equation (4.5) we obtain:

(4.9)

d_2 = \frac{n}{2 - n}d'.

Now using the exact geometric relations:

(4.10) $\delta = 2\alpha - 2\beta$

and the Snell’s law (4.4) in a small-angle limit, we finally express $d_2$ in terms of $d$ :

(4.11)

d_2 = \frac{n}{9n - 10}d,

and the velocity $v_2$ of the second image in terms of $v$ :

(4.12)

v_2 = \frac{n}{9n - 10}v = \frac{2}{3}v.

The relative velocity of the two images is:

(4.13) $\mathbf{v}_{\mathrm{rel}} = \mathbf{v}_1 - \mathbf{v}_2$

in a vector form. Since vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ are oppositely directed (one of the images moves upward, the other, downward) the magnitude of the relative velocity is:

(4.14)

v_{\mathrm{rel}} = v_1 + v_2 = \frac{8}{3}v.

Оригинал в Архива: ipho71_problems.pdf · решение от архива: ipho71_solutions.pdf · допълнителни данни: 1971_Sofia_Problems.pdf · допълнителни данни: 1971_Sofia_Solutions.pdf