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For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000Å and electrons accelerated through 100 V used as the illuminating substance. |
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Answer» Solution :Formula of linear limit of resolution of compound MICROSCOPE having oil immersion objective is : `d_(m)=(1.22lamda)/(2nsinbeta)` Now resolution power, `P=(1)/(d_(m))=(2nsinbeta)/(1.22lamda)` `:.Pprop(1)/(lamda)` `:.(P_(1))/(p_(2))=(lamda_(2))/(lamda_(1))""......(1)` Here, in the second case `W=DeltaK` `:.Ve=(1)/(2)mv^(2)=(P^(2))/(2M)""(":.P=mv)` `:.Ve=(h^(2))/(2mlamda^(2))` ( `"":.P=(h)/(lamda)` h=Planck.s CONSTANT) `:.lamda^(2)=(h^(2))/(2Vem)` `:.lamda=(h)/(SQRT(2Vem))` `:.lamda_(2)=(h)/(sqrt(2Vem))""(":.` In the second case `lamda=lamda_(2)`) `:.lamda_(2)=(6.625xx10^(-34))/(sqrt(2xx100xx1.6xx10^(-19)xx9.1xx10^(-31)))` `:.lamda_(2)=(6.625xx10^(-9))/(sqrt(200xx1.6xx9.1))m` `:.lamda_(2)=(6.625xx10^(-9))/(sqrt(200xx1.6xx9.1))xx10^(10)Å` `=(66.25)/(sqrt(200xx1.6xx9.1))` `=(66.25)/(53.96)` `:.lamda_(2)=1.2278Å` Now from equation (1), `(P_(1))/(P_(2))=(1.2278)/(5000)` `:.(P_(1))/(P_(2))=2.456xx10^(-4)` |
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