
#1
Apr605, 12:51 PM

P: 1,445

A physicist wishes to deposit a thin film on both sides of a glass (with index of refraction 1.50) window to reduce reflected light of wavelength 532nm to 0%
In order to achieve 100% destructive interference for light reflectedo ff of the air /thing film inteference and the light reflected off the thin film/glass interface what must hte index of refraction of the thin film be?? so its is purely destructive [tex] 2d + \frac{1}{2} \lambda = (m+ \frac{1}{2}) \frac{\lambda}{n} [/tex] what is the value of m though?? So would i assume m=0?? But then what is the value of the thickness?? Sincei want to find the index of refraction ,n. Perhaps the subsequent sub questions have somethin to do with this answer? Sketch a graph of the reflected intensity for lambda = 532nm as a function of thickness T 0<T<2000 nm I could do this if i knew the index of refraction... What thickness of a film will give this 100% AR effect at 532nm? What fraction of light at twice the laser wavelength lambda = 1264nm will be reflected from this AR coated window? P.S. Dont ask me what AR is, i dont know either! 



#2
Apr605, 05:21 PM

P: 1,445

can anyone help??




#3
Apr605, 05:41 PM

P: 2,223

Just wondering what level physics is this? course title i mean




#4
Apr605, 07:58 PM

Mentor
P: 40,889

Thin film interference
AR = AntiReflection
There are several things to consider: (1) Interference: You want the light reflected off the first and second surfaces to destructively interfere. This happens when the thickness (d) meets this criteria: [itex]2dn/\lambda = k + 1/2[/itex], where k = 0, 1, 2... The thinnest coating meeting this criteria will be when k = 0. Of course, you need to know what n is for the material of the coating. (In this case you know it will be between 1 and 1.5.) (2) Reflectance: To have perfect antireflection you must choose the index of refraction of the coating so that the intensity of both reflections is equal. Thus you want the reflectances at each surface to be equal. (Reflectance depends on the indices of refraction of the surfaces involved: look it up!) 



#5
Apr605, 09:10 PM

P: 1,445

It is supposed to be part two of the EM course incorporating its applications to optics (such as maxwell's equations) 



#6
Apr605, 09:24 PM

P: 1,445

I do not know the significance of the reflection coefficinet does it give the fracton of intensity reflected?? in [itex]2dn/\lambda = k[/itex] we known lambda and we know k we dont know n and d but we cant find n as a function of d. So any thickness would give a reasonable value for n?? I dont think guessing is what im supposed ot be doing here! 



#7
Apr705, 08:00 AM

Mentor
P: 40,889




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