InterviewSolution
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InterviewSolution
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A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young's double-slit experiment. (a) Find the distance of the third bright fringe on the screen, from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelength coincide ? |
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Answer» Solution :Here `lamda_(1)=650nm=6.5xx10^(-7)m,lamda_(2)=520nm=5.2xx10^(-7)m,d=1mm=10^(-3)m,D=60cm=0.6m` (a) Distance of third bright fringe from the central bright fringe for wavelength `lamda_(1)=650nm` `x_(3)=(3Dlamda)/(2)=(3xx0.6xx6.5xx10^(-7))/(1xx10^(-3))=1.17xx10^(-3)m=1.17mm` (b) Let at a distance x from central MAXIMA the bright fringes due to wavelength `lamda_(1) and lamda_(2)` coincide first time. for this to happen, `nlamda_(1)=(n+1)lamda_(2)`, where n is an integer `IMPLIES(n+1)/(n)=(lamda_(1))/(lamda_(2))=(6.5xx10^(-7))/(5.2xx10^(-7)m)=(5)/(4)impliesn=4` It means that at distance nth (4th) maxima for wavelength `lamda_(1)` is just coinciding with (n+1)st (5TH) maxima for wavelength `lamda_(2)` `therefore=x=(nDlamda_(1))/(d)=(4xx0.6xx6.5xx10^(-7))/(1xx10^(3))=1.56xx10^(-3) m=1.56mm`. |
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