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In Young's double slit experiment, the distance between sources is $1 \mathrm{~mm}$ and distance between the screen and source is $1 \mathrm{~m}$. If the fringe width on the screen is $0.06 \mathrm{~cm}$, then $\lambda=$
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$6000 Å$
$\beta=\frac{\lambda D}{d} \Rightarrow\left(0.06 \times 10^{-2}\right)=\frac{\lambda \times 1}{1 \times 10^{-3}} \Rightarrow \lambda=6000 Å$
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