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In Lloyd's single mirror experiment A light wave emitted directly from sources s interferes with a wave reflected from a mirror M. The interference pattern is formed on the screen which is kept perpendicular to the mirror and at a distance. $D=1 m$ from the source as shown . If the sources is at a distance of 1 mm from the plane of the mirror what is the fringe width when monochromatic light of 600 nm is incident from the sources




$(a)\;6 \times 10^{-7} \;m \\ (b)\;3 \times 10^{-7} \;m \\ (c)\;4 \times 10^{-6}\;m \\ (d)\;3 \times 10^{-6} $

1 Answer

Fringe width $= \large\frac{D \lambda}{d}$
Where d - is separation between two sources.
Since $S_2 $ is image of $S_1$
$S_1 S_2 =2 \times 1 m =2m$
$\therefore \beta =\large\frac{600 \times 10^{-9} \times 1}{2}$
$\qquad= 300 \times 10^{-9} \;m$
$\qquad= 3 \times 10^{-7} \;m$
Hence b is the correct answer.
answered Jan 24, 2014 by meena.p

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