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In YDSE $\large\frac{d}{D}$$=10^4$ (d= distance between slits ; D= distance of screen from slits). at a point P on screen resulting intensity is equal to intensity due to individual slit $I_0$ Then the distance of point P from central maximum is $(\lambda= 6000 A^{\circ})$

$(a)\;2 \;mm \\ (b)\;0.5\;mm \\ (c)\;1\;mm \\ (d)\;4\;mm $

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1 Answer

$I= 4I_0 \cos ^{2} \bigg( \large\frac{\phi}{2} \bigg)$
$I_0 =4 I_0 \cos ^2 \bigg( \large\frac{\phi}{2} \bigg)$
$\therefore \cos \bigg(\large\frac{\phi}{2} \bigg)=\large\frac{1}{2}$
or $\large\frac{\phi}{2}=\frac{\pi}{3}$
or $\phi =\large\frac{2 \pi}{3}= \bigg( \large\frac{2 \pi}{\lambda}\bigg).$$ \Delta x$
or $\large\frac{1}{3} =\bigg( \large\frac{1}{\lambda}\bigg) .y. \large\frac{d}{D}$
$( \Delta x =y. \large\frac{d}{D}\bigg)$
$\therefore y= \large\frac{\lambda}{3 \times \Large\frac{d}{D} } $
$\qquad=\large\frac{ 6 \times 10^{-7}}{3 \times 10^{-4}}$
$\qquad= 2 \times 10^{-3} m$
$\qquad= 2 \;mm$
Hence a is the correct answer.
answered Feb 19, 2014 by meena.p

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