На дифракционную решетку с периодом 0,001мм падает белый свет. На экране, находящемся на расстоянии 1,5м от решетки образуются картина дифракции света. Определите расстояние на экране между первым и вторым максимумом фиолетового света λ=400нм

На дифракционную решетку с периодом 0,001мм падает белый свет. На экране, находящемся на расстоянии 1,5м от решетки образуются картина дифракции света. Определите расстояние на экране между первым и вторым максимумом фиолетового света λ=400нм
- Предмет:
  Физика
- Автор:
  miss piggyqomg
- 6 лет назад

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Пусть a- расстояние от центрального максимума до данного. При малых углах sina=tga=a/S S=1,5 м d=10^-6 мd*tga=k*Ld*a/S=k*La=k*L*S/d a1=k1*L*S/d a2=k2*L*S/dda=a2-a1=L*S*(k2-k1)/d=400*10^-9*1,5*(2-1)/10^-6=0,6 м
- Автор:
  mcmahon
- 6 лет назад
- 0
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