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In Young's double slit experiment , two wavelengths lamda_1 = 780 nmand lamda_2 = 520 nm are used to obtain interference fringes. If the n^(th) bright band due to lamda_1 coincides with (n+1)^(th) bright band due to lamda_2, then the value of n is

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Solution :`x_(N)=x_(n+1)`
`""^(n)beta_(1)=(n+1)beta_(2)`
`(nlamda_(1)D)/d=(n+1)(lamda_(2)D)/d`
`nlamda_(1)=(n+1)lamda_(2)`
`(n+1)/n=lamda_(1)/lamda_(2)=780/520`
`52n+52=78n`
26n = 52
n = `52/26` = 2


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