12_Ch 24 College Physics ProblemCH24 Wave Optics

# 12_Ch 24 College Physics ProblemCH24 Wave Optics - 314.20...

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Unformatted text preview: 314 CHAPTER 24 24.20 The transmitted light is brightest when the reflected light is a minimum (that is, the same conditions that produce destructive interference in the reflected light will produce constructive interference in the transmitted light). As light enters the air layer from glass, any light reflected at this surface has zero phase change. Light reflected from the other surface of the air layer (where light is going from air into glass) does have a phase reversal. Thus, the condition for destructive interference in the light reflected from the air film is 2 t = m λn , m = 0 , 1, 2,… Since λn = λ n film condition is 24.21 = λ 1.00 = λ , the minimum non-zero plate separation satisfying this d = t = ( 1) λ 2 = 580 nm = 290 nm 2 (a) For maximum transmission, we want destructive interference in the light reflected from the front and back surfaces of the film. If the surrounding glass has refractive index greater than 1.378, light reflected from the front surface suffers no phase reversal, and light reflected from the back does undergo phase reversal. This effect by itself would produce destructive interference, so we want the distance down and back to be one whole wavelength in the film. Thus, we require that 2t = λn = λ n film or t= λ 2 n film = 656.3 nm = 238 nm 2 ( 1.378 ) (b) The filter will expand. As t increases in 2n film t = λ , so does λ increase (c) Destructive interference for reflected light happens also for λ in 2 t = 2 λ n film , or λ = n filmt = ( 1.378 ) ( 238 nm ) = 328 nm (near ultraviolet) ...
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