Вероятно, это уж слишком, но я не могу устоять перед искушением привести вам еще один интересный пример того, как ведет себя свет и как можно проанализировать его поведение, пользуясь правилами последовательных этапов. Поместим детектор под стеклом и рассмотрим вопрос, о котором мы не говорили в первой лекции, – о вероятности прохождения света через две поверхности стекла (см. рис. 43).
Вы, конечно, знаете ответ: вероятность того, что фотон попадет в В, равна просто 100 % за вычетом найденной раньше вероятности, что фотон попадет в А. Так, если мы получили, что вероятность попасть в А равна 7 %, то вероятность попасть в В равна 93 %. А так как вероятность для А меняется от 0 через 8 % до 16 % (для разных толщин стекла), то вероятность для В меняется от 100 % через 92 % до 84 %.
Это правильный ответ, но мы ожидаем, что все вероятности можно вычислять, возведя в квадрат результирующую стрелку. Как вычислить амплитуду пропускания света стеклянной пластинкой? И как ей удается таким именно образом менять свою длину, чтобы всегда соответствовать длине амплитуды А, так что вероятность для А и вероятность для В в сумме всегда дают 100 %? Давайте рассмотрим вопрос несколько подробнее.
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Рис. 43. Прохождение через две поверхности можно разделить на пять этапов. На этапе 2 единичная стрелка сжимается до 0,98, на этапе 4 стрел-ка длиной 0,98 сжимается еще раз до 0,98 (что дает примерно 0,96). На этапах 1, 3 и 5 происходит только поворот. В итоге квадрат стрелки длиной 0,96 будет равен примерно 0,92, что дает вероятность прохождения через две поверхности, равную 92 %. (Это соответствует 8 %-ному отражению, правильному лишь «дважды в сутки».) Когда толщина пластинки такова, что вероятность отражения равна 16 %, то в сумме с 92 % вероятности прохождения получается 108 %, т. е. мы учли 108 % света. В нашем анализе что-то неправильно!
Движение фотона от источника к детектору, находящемуся под стеклом, состоит из пяти этапов. Давайте сжимать и поворачивать единичную стрелку по мере продвижения.
Первые три этапа будут такими же, как в предыдущем примере: фотон летит из источника к стеклу (поворот, сжатия нет), фотон проходит сквозь переднюю поверхность (поворота нет, сжатие до 0,98); фотон проходит стекло (поворот, сжатия нет).
Четвертый этап – когда фотон проходит сквозь заднюю поверхность стекла – ничем не отличается от второго этапа в том, что касается поворотов и сжатия: поворота нет, а сжатие до 0,98 от 0,98, т. е. длина стрелки становится 0,96.
Наконец, пятый этап – фотон опять летит по воздуху в детектор – это значит, что происходит еще поворот, но без дальнейшего сжатия. В результате получаем стрелку длиной 0,96, указывающую в некотором направлении, заданном последовательными поворотами часовой стрелки.