InterviewSolution
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InterviewSolution
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In a young ,s double slit experiment distance the slits is 1mm. The fringe width is found to be 0.mm. when the screen is moved through a distance of 0.25m away from the plane of the slit used . |
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Answer» Solution : `BETA= 0.6xx10^(-3)m,beta_(2)=0.75xx10^(-3)m` `d= 1xx10^(-3)m, D_(1) = Dm, D_(2) = (D+ 0.25 ) m, lamda= ?` `beta_(1) = (lamdaD_(1))/(d) = (lamdaD)/(1xx10^(-3)) rightarrow(1)` `beta_(2)= ( lamda(D_(2)))/(d) = (lamda(D+0.25))/(1xx10^(-3)) rightarrow(2)` `beta-(1))/(beta_(2))= (0.6xx10^(-3))/(0.75xx10^(-3))= (lamdaD)/(lamda(D + 0.25))` `(D+ 0.25)0.6xx10^(-3)= 0.75xx10^(-3)D` `RightarrowD=1m` Subsituting the value of D in eqaution (1) or (2) `lamda= (beta_(1)d)/(D)= (0.6xx10^(-3)xx10^(-3))/(1)` `lamda=6xx10^(-7)m= 6000A^(@)` |
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