$A$ plate of thickness $t$ made of a material of refractive index $\mu$ is placed in front of one of the slits in a double slit experiment. What should be the minimum thickness $t$ for which the intensity at the centre of the fringe pattern is zero?

In the $YDSE$ shown,the two slits are covered with thin sheets having thickness $t$ and $2t$,and refractive index $2\mu$ and $\mu$ respectively. Find the position $(y)$ of the central maxima.

The figure shows a double-slit experiment. Each slit has a width $W$. $A$ thin glass slab of thickness $\delta$ and refractive index $\mu$ is placed in front of one of the slits. The intensity at the central point $C$ is measured as a function of the thickness $\delta$. At what thickness $\delta$ will the intensity at $C$ be minimum?

In a Young's double-slit experiment,a glass slab of thickness $1.2 \, \mu m$ and refractive index $1.5$ is placed in front of one slit. Another slab of thickness $t$ and refractive index $2.5$ is placed in front of the other slit. If the position of the central fringe remains unchanged,then the thickness $t$ is equal to $...... \, \mu m$.

In a Young's double-slit experiment,light of wavelength $4800 \, \mathring{A}$ is used. One slit is covered by a thin transparent plate of refractive index $1.4$ and the other slit is covered by another thin transparent plate of refractive index $1.7$. As a result,the central bright fringe shifts to the position where the fifth bright fringe was previously formed. If the thickness of both plates is the same,find the thickness in $\mu m$.

The minimum thickness of a mica sheet having refractive index $\mu = 3/2$,which should be placed in front of one of the slits in $YDSE$ to reduce the intensity at the center of the screen to half of the maximum intensity,is: