IYPT 1989, Задачи на турнира — Задача 14. Fence

Problems for the 2nd International Young Physicists’ Tournament

Critical edition

Olympiets Youth Center, Moscow, Soviet Union; March 24‒April 2, 1988 [1]

Critical edition: translated, restored, and commented text [2]

FINAL DRAFT. — Please do not re-publish. Suggestions and criticism welcome

Translated, edited and commented by Ilya Martchenko. The original edition’s historical notes, sources and acknowledgments are retained with problem 1.

Епиграф

Squamps are hunted from the inside only.
S. Lem [3]

Условие

A remote large object is separated from you by a picket fence. It happens that you can see the object if you do not stay near the fence, but go along the fence in a car. Explain this phenomenon. What speed is sufficient if $a$ is width of a fence board, $b$ is width of the gaps, $L$ is distance to the fence ($L\gg a,b$), $\gamma$ is angular size of the remote object, $\gamma\gg\frac{a+b}{L}$. [27]

Note [26]

A different problem at the Correspondence Round [Kvant 1988]: “Triboluminescence. How much sugar would you need to read a message (a few words) from your predecessor in a dark dungeon? You are out of matches and you have already thrown away your lighting.” In [Jordens 1989], the problem reads: “Triboluminescence. How much sugar do you need to read a message (several words) from your predecessor in an underground vault? You have no matches and you have thrown out the lantern”

Note [27]

The 1990 Czech text reads “$\gamma\gg\frac{a+b}{L}$”, but the Slovak text relying on this Czech source reads “$\gamma\ll\frac{a+b}{L}$”. The Hungarian version states “$\gamma>>(a+b)/L$”. The 1996 Czech and the 2006 English texts do not introduce the variable $\gamma$ and pose any conditions, but just say “the angular size of the remote object is $\left(\frac{a+b}{L}\right)$” [Kluiber 1996], “the angle size of the remote object is ((a+b)/L)” [Proceedings 2006]

Решение

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Оригинал в Архива: 1989.pdf