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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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A thin film of oil of $\mu =1.2 $ floats on watch of $\mu_w =\large\frac{4}{3}$. A monochromatic light of wave length $\lambda= 9600 A^{\circ} $ falls normally and it appears dark. What must be minimum change in thickness of oil film which will appear bright if normally reflected by same light.

$(a)\;10^{-7}\;m \\ (b)\;2 \times 10^{-7}\;m \\ (c)\;3 \times 10^{-7} \\ (d)\;5 \times 10^{-7}\;m $

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1 Answer

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The path difference between two reflected light from upper and lower surface of oil film is $2 \mu t$
Now for dark fringe to turn to bright fringe the change in the path difference should be minimum $\large\frac{ \lambda}{2}$
If $\Delta t$ is change in thickness
$2 \mu (\Delta t) =\large\frac{\lambda}{2}$
$\Delta t =\large\frac{\lambda}{4 \mu}$
$\qquad= \large\frac{9.6 \times 10^{-7}}{4 \times 1.2}$
$\qquad= 2 \times 10^{-7}\; m$
Hence b is the correct answer.
answered Jan 23, 2014 by meena.p
 

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