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In a young's double slit experiment the fringe width is $0.05\;mm$ when the distance of screen from slit is $75 \;cm$ . If the distance of screen from the slit is increased by $25 \;cm$ , the fringe width will be

$(a)\;0.66\;mm \\ (b)\;0.066\;mm \\ (c)\;6.0\;mm \\ (d)\;60\;mm $

1 Answer

$\large\frac{\beta '}{\beta} =\large\frac{\Large\frac{\lambda D'}{d}}{\Large\frac{\lambda D}{d}}$
$\qquad= \large\frac{D'}{D}$
$D= 75$
$D'= 75+25=100$
$\large\frac{\beta'}{\beta}= \large\frac{100}{75}$
$\beta' =0.05 \times \large\frac{4}{3}\; $$mm$
$\qquad=0.066 \;mm$
Hence b is the correct answer.
answered Jan 23, 2014 by meena.p

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