Minimum thickness of a mica sheet having mu = | vedclass

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(C) The intensity at any point in $YDSE$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_{\max} = 4I_0$ and $\phi$ is the phase difference.Given that t

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$\lambda /2$

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(C) The intensity at any point in $YDSE$ is given by $I = 4I_0 \cos^2(\phi/2)$,where $I_{\max} = 4I_0$ and $\phi$ is the phase difference.Given that the intensity at the center becomes half of the maximum intensity,we have $I = I_{\max}/2 = 2I_0$.Substituting this into the formula: $2I_0 = 4I_0 \cos^2(\phi/2) \Rightarrow \cos^2(\phi/2) = 1/2 \Rightarrow \phi/2 = \pi/4 \Rightarrow \phi = \pi/2$.The phase difference introduced by the mica sheet of thickness $t$ and refractive index $\mu$ is $\phi = (2\pi/\lambda)(\mu - 1)t$.Equating the two expressions for $\phi$: $\pi/2 = (2\pi/\lambda)(3/2 - 1)t$.$\pi/2 = (2\pi/\lambda)(1/2)t = \pi t / \lambda$.Solving for $t$,we get $t = \lambda/2$.

Minimum thickness of a mica sheet having $\mu = 3/2$ which should be placed in front of one of the slits in $YDSE$ is required to reduce the intensity at the centre of the screen to half of the maximum intensity is-

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