奇摩浪客 asked in 科學其他:科學 · 1 decade ago

請物理電磁學高手指教,反射及折射,20點,謝謝!

假設為正入射,證明有可能可以排除掉光的反射的一個波長當進入及離開一個玻璃片時、靠覆蓋一層厚度為λ/4的電介質〈誘導性的、非傳導性的,不知翻那個,反正是dielectric〉薄膜在其表面上,此λ為薄膜內的光波長。請解釋這個特定的厚度的重要性。〈提示:建立四個波、入射、薄膜前面、薄膜後面及傳至玻璃裡。小心考量薄膜裡的相移。〉

判定最理想的折射率及薄膜的厚度當玻璃的折射率為1.5,運作550nm的波長於真空中。

我怕翻錯,所以將原文複製如下:

Assuming normal incidence, show that it is possible to eliminate the reflection of light of one wavelength leaving air and entering a glass plate of reflective index ng by coating the surface with a film of dielectric whose thickness is λ/4, whereλ is the wavelength of the light in the film. Explain the significance of this particular thickness.

(Hint: set-up 4 waves, incident, forward in film, backward in film and transmitted in glass. Consider carefully phase changes with the film.)

Determine the optimal refractive index and thickness of the film to be applied of a glass of refractive index 1.5, when operating at a wavelength in vacuo of 550 nm.

Update:

如果可以的話,請至我的發問區看看,還有一些疑問,萬分感謝!

Update 2:

我知道有點長,20點好像很小,不過我還有其他一些很簡單的問題,也都是20點,有些還滿好玩的!

1 Answer

Rating
  • Anonymous
    1 decade ago
    Favorite Answer

    請見下圖:

    圖片參考:http://i117.photobucket.com/albums/o61/billy_hywun...

    首先, 薄膜為一相對空氣較光密的介質, 因此, 在 1 → 2 的反射中, 光線有一個 π 的相位轉移 (phase change π), 而在 3 → 4 的反射中亦有發生 (因為在它之下的玻璃亦相對它為較光密的介質). 因此 2 和 5 兩波是同相 (In phase) 的.

    所以, 2 和 5 兩波的光學程差 (optical path difference) 為 2ngt, 其中 ng 為薄膜的折射率. 而此兩波若要發生相消干涉 (Destructive interference), 則此程差必需是至少為薄膜中的波長的 1/2 倍.

    所以:

    2t = λ/2 (由於是 normal incidence, 所以事實上的程差為薄膜的兩倍)

    t = λ/4

    因此結果便驗證了.

    若光波在真空的波長為 550 nm, 它在薄膜內的波長為 550/1.5 = 367.7 nm

    所以此時薄膜的厚度應為 367.7/4 = 91.7 nm

    Source(s): My physics knowledge
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