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In Young's experiment fifth bright fringe produced by light of 4000Åsuperposes on the fourth bright fringe of an unknown wavelength. Find the unknown wavelength. |
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Answer» Solution :Here, `lamda_(1)=4000Å,n_(1)=5,n_(2)=4` Distance between two consecutive fringes for wavelength `lamda_(1)` is `beta_(1)` and that for wavelength `lamda_(2)` it is `beta_(2)`. Here distance of `n_(1)^(TH)` fringe from central MAXIMUM=distance of `n_(2)^(th)` fringe from central maximum. `:.n_(1)beta_(1)=n_(2)beta_(2)` `:.n_(1)(lamda_(1)D)/(d)=n_(2)(lamda_(2)D)/(d)""[:.beta=(lamdaD)/(d)]` `:.n_(1)lamda_(1)=n_(2)lamda_(2)` `:.lamda_(2)=(n_(1)lamda_(1))/(n_(2))` `:.lamda_(2)=(n_(1)lamda_(1))/(n_(2))` `:.lamda_(2)=(5xx4000)/(4)` `:.lamda_(2)=5000Å` |
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