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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is
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five
five
For interference maxima, $d \sin \theta=n \lambda$
Here $d=2 \lambda$
$\therefore \sin \theta=\mathrm{n} / 2$ and is satisfied by 5 integral values of $\mathrm{n}(-2,-1,0,1,2)$, as the maximum value of $\sin \theta$ can only be 1 .
Here $d=2 \lambda$
$\therefore \sin \theta=\mathrm{n} / 2$ and is satisfied by 5 integral values of $\mathrm{n}(-2,-1,0,1,2)$, as the maximum value of $\sin \theta$ can only be 1 .
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