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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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In a Young's double slit experiment separation between two consecutive dark. fringes is $1.2 \;mm$, wave length $\lambda$ of light is $600 \;nm$ Distance between source and screen $D= 1 m $, the separation between slits is

$(a)\;1\;mm \\ (b)\;0.05\;mm \\ (c)\;0.5 \;mm \\ (d)0.1 \;mm $

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1 Answer

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Fringe width $\beta= \large\frac{D \lambda}{d}$
'd' separation between slits
$ 1.2 \times 10^{-2} =\large\frac{1 \times 60 \times 10^{-10}}{d}$
$d= \large\frac{600 \times 10^{-10}}{1.2 \times 10^{-3}}$
$\quad= 0.5 mm$
Hence c is the correct answer.
answered Jan 21, 2014 by meena.p
 

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