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InterviewSolution
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Find the minimum thickness of a film of refractive index 1.25, which will storngly reffect the light of wavelenght 589 nm. Also find the minimum thikness of the film to be anti - reflecting. |
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Answer» Solution :`lambda = 589 nm = 589 xx 10^(-9) m` For the film to have strong reflection, the reflected waves should INTERFERE constructively The least optical path difference introduced by the film should be `lambda//2` . The optical path difference between the waves reflected from the two surface of the film is 2pd. Thus, for strong reflection, 2ND = `lambda//2(N - 1)` Rewriting, ` d = (lambda)/(4mu)` Substituting, `d=(589xx10^(9))/(4xx1.25)=117.8xx10^(-9)` `d = 117.8 xx 10^(-9) = 117.8 nm` For the film to be anti - reflecting, the reflected rays should interfere destructively. The least optica path difference introduced by the film should be `lambda`. The optical path difference between the waves reflected from the surface of the film is 2pd. For strong reflection. `2mud = lambda (n = 1)`. Rewriting, `d = (lambda)/(2pi)` Substituting , `d=(589xx10^(9))/(2xx1.25)=235.6xx10^(-9)` `d = 235.6 xx 10^(-9) = 235.6 nm` |
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