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In a YDSE, $D = 1m \; d= 1mm$ and $\lambda =\large\frac{1}{2}$$mm$ Then the distance between the first and central maxima on screen

$(a)\;\frac{1}{2 \sqrt 3}\;m \\ (b)\;\frac{1}{\sqrt 3} m \\ (c)\;\frac{1}{\sqrt 2}m \\ (d)\;\frac{2}{\sqrt 3}\;m $

1 Answer

$D >> d$
Hence the path difference $\Delta P= d \sin \theta$
Clearly $, n << \large\frac{d}{\lambda}$$=2$
It is not possible for any values of n .
Hence $\Delta P=\large\frac{dy}{D} $ can not be used
for 1st maxima,
$\Delta P= d \sin \theta=\lambda$
$=> \sin \theta =\large\frac{\lambda}{D} =\frac{1}{2}$
$=> \theta= 30^{\circ}$
Hence $y= D \tan \theta =\large\frac{1}{\sqrt 3}$$m$
Hence b is the correct answer.


answered Feb 13, 2014 by meena.p
edited Jul 22, 2014 by thagee.vedartham

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