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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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Find the maxima and minima obtained on screen in a $YDSE, D= 1m,d=1mm$ and $ \lambda=\large\frac{1}{2}\;mm$

$(a)\;(4,5) \\ (b)\;(5,4) \\ (c)\;(3,4) \\ (d)\;(4,3) $

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1 Answer

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Maximum path difference $\Delta P_{max}=d= 1mm$
=> highest order maxima,
$n_{max} =\bigg[\large\frac{d}{\lambda} \bigg]$$=2$
and highest order minima $n_{min}= \bigg[ \large\frac{d}{\lambda} +\frac{1}{2}\bigg]$
$\qquad= 2$
Total number of maxima $=2 n _{max} +1=5$
Total number of minima $=2n _{min}=4$
Hence the correct answer is b
answered Feb 13, 2014 by meena.p
 

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