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# In a YDSE, $D = 1m \; d= 1mm$ and $\lambda =\large\frac{1}{2}$$mm Then the distance between the first and central maxima on screen (a)\;\frac{1}{2 \sqrt 3}\;m \\ (b)\;\frac{1}{\sqrt 3} m \\ (c)\;\frac{1}{\sqrt 2}m \\ (d)\;\frac{2}{\sqrt 3}\;m Can you answer this question? ## 1 Answer 0 votes D >> d Hence the path difference \Delta P= d \sin \theta \large\frac{d}{\lambda}$$=2$
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.

edited Jul 22, 2014