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In a biprism experiment, a slit is illuminated by a light of wavelength `4800`. The distance between the slit and biprism is and the distance between the biprism is `15cm` and the distance between the biprism and eyepiece is `85cm`. If the distance between virtual sources is `3mm`, determine the distance between `4th` bright band on one side and `4th` dark band on the other side of the central bright band. |
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Answer» Given : ` lambda =4800=4800xx10^(-10)m` `D=85+15=100cm=1m` `d=3mm=3xx10^(-3)m` Distance of `4th` bright band on one side `=(4Dlambda)/(d)` `=(4xx1xx4800xx10^(-10))/(3xx10^(-3))` `=6.4xx10^(-4)m` `4th` dark band on the other side `=(m-(1)/(2))(Dlambda)/(d)` `=((4-(1)/(2))xx4800xx10^(-10)xx1)/(3xx10^(-3))` `=0.00056` `=5.6xx10^(-4)m` `:.` Total distance `=5.6xx10^(-4)+6.4xx10^(-4)` `=1.2xx10^(-3)m`. |
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