1.

A fringe width of a certain interferences pattern is beta = 0.002 cm. What is distance of 5th dark fringe from centre ?

Answer»

`9 xx 10^(-3) CM`
`11 xx 10^(-2) cm`
`1.1 xx 10^(-2) cm`
`3.28 xx 10^(6) cm`

Solution :The distance of 5m DARK fringe from centre is given by :
`x_(n) = (2N + 1)(lambdal)/(2d)` as n = 4 for 5th dark fringe
So, `x_(5) = (9)/(2) (lambdaD)/(d)`
as `(lambdaD)/(d) = beta`
So, `x_(5) = (9/2)beta = 9/2 xx 0.002`
` = 9 xx 10^(-3) cm`.


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