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In an young double slit experiment the slit separation is 3 times the wave length of the light used What is the maximum number of interference maxima possible

(a) infinite




1 Answer

If d is slit separation ; $d= 3 \lambda$
We have for maxima
$d \sin \theta = n \lambda$
$\sin \theta =\large\frac{n \lambda}{d}$
$\sin \theta$ can take maximum value of 1
$\large\frac{ n \lambda }{d}$$ \leq 1$
$n \leq \large\frac{d}{\lambda}$
$ n \leq \large\frac{3 \lambda}{\lambda}$
$n \leq 3$
So we can have 3 maximas on either side of central maxima ie 7 maximas.
Hence d is the correct answer
answered Jan 22, 2014 by meena.p

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