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A fringe width of a certain interferences pattern is beta = 0.002 cm. What is distance of 5th dark fringe from centre ? |
Answer» <html><body><p>`9 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>`<br/>`11 xx 10^(-2) cm`<br/>`1.1 xx 10^(-2) cm`<br/>`3.28 xx 10^(6) cm`</p>Solution :The distance of 5m <a href="https://interviewquestions.tuteehub.com/tag/dark-943775" style="font-weight:bold;" target="_blank" title="Click to know more about DARK">DARK</a> fringe from centre is given by : <br/> `x_(n) = (<a href="https://interviewquestions.tuteehub.com/tag/2n-300431" style="font-weight:bold;" target="_blank" title="Click to know more about 2N">2N</a> + 1)(lambdal)/(2d)` as n = 4 for 5th dark fringe <br/> So, `x_(5) = (9)/(2) (lambdaD)/(d)` <br/> as `(lambdaD)/(d) = beta` <br/> So, `x_(5) = (9/2)beta = 9/2 xx 0.002` <br/> ` = 9 xx 10^(-3) cm`.</body></html> | |