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Консультация № 177472
25.03.2010, 22:31
0.00 руб.
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1
Здравствуйте уважаемые эксперты, пожалуйста помогите решить вот эти задачи 
:
1)Луч света падает на плоскопараллельную стеклянную пластину показателем преломления n=1,6 под углом φ = 45 градусов. Определить толщину d пластины, если смещение вышедшего из пластины луча составляет h= 2 см.
2)Для уменьшения доли отраженного света от поверхности стекла на нее наносят тонкую пленку, показатель преломления которой меньше показателя преломления стекла. Какой наименьший толщины dmin пленку с показателем преломления n = 4/3 надо нанести на поверхность стекла, чтобы при падении (нормально к поверхности) света, содержащего излучение двух длин волн лямбда 1 = 700 нм и лямбда 2 = 420 нм, отраженный свет был максимально ослаблен для обеих длин волн?
:1)Луч света падает на плоскопараллельную стеклянную пластину показателем преломления n=1,6 под углом φ = 45 градусов. Определить толщину d пластины, если смещение вышедшего из пластины луча составляет h= 2 см.
2)Для уменьшения доли отраженного света от поверхности стекла на нее наносят тонкую пленку, показатель преломления которой меньше показателя преломления стекла. Какой наименьший толщины dmin пленку с показателем преломления n = 4/3 надо нанести на поверхность стекла, чтобы при падении (нормально к поверхности) света, содержащего излучение двух длин волн лямбда 1 = 700 нм и лямбда 2 = 420 нм, отраженный свет был максимально ослаблен для обеих длин волн?
Обсуждение
26.03.2010, 12:04
общий
это ответ
Здравствуйте, Vika_88.
Решение 2 задачи:
Условие минимума - дельта=(2k+1)(лямбда/2)
Из чертежа видно, что оптическая разность хода
дельта=l2n2-l1n1=(AB+BC)n2-ADn1=(2k+1)(лямбда/2)
В нашем случае получаем:
дельта=2dn2-(2k+1)(лямбда/2)
Откуда искомая толщина пленки
d=((2k+1)лямбда)/(4*пи)
По условию задачи минимум должен быть для обеих длин волн, поэтому:
d=((2k1+1)лямбда1)/(4*пи)
d=((2k2+1)лямбда2)/(4*пи)
Левые части равны, приравниваем правые части - получаем:
((2k1+1)лямбда1)/(4*пи)=((2k2+1)лямбда2)/(4*пи)
(2k1+1)*лямбда1=(2k2+1)*лямбда2
Подставив числовые данные получим, что k1=1, а k2=2
Вычислим искомую величину d:
d=(2*1+1)*700*10^-9/(4*3,14)=0,17*10^-6(м)=0,17(мкм)
Решение 2 задачи:
Условие минимума - дельта=(2k+1)(лямбда/2)
Из чертежа видно, что оптическая разность хода
дельта=l2n2-l1n1=(AB+BC)n2-ADn1=(2k+1)(лямбда/2)
В нашем случае получаем:
дельта=2dn2-(2k+1)(лямбда/2)
Откуда искомая толщина пленки
d=((2k+1)лямбда)/(4*пи)
По условию задачи минимум должен быть для обеих длин волн, поэтому:
d=((2k1+1)лямбда1)/(4*пи)
d=((2k2+1)лямбда2)/(4*пи)
Левые части равны, приравниваем правые части - получаем:
((2k1+1)лямбда1)/(4*пи)=((2k2+1)лямбда2)/(4*пи)
(2k1+1)*лямбда1=(2k2+1)*лямбда2
Подставив числовые данные получим, что k1=1, а k2=2
Вычислим искомую величину d:
d=(2*1+1)*700*10^-9/(4*3,14)=0,17*10^-6(м)=0,17(мкм)
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