# 請物理電磁學高手指教，反射及折射，20點，謝謝！

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:

Update 2:

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• Anonymous

請見下圖：

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