In a Young's double slit experimental arrangement shown here,if a mica sheet of thickness $t$ and refractive index $\mu$ is placed in front of the slit $S_1$,then the path difference $(S_1P - S_2P)$

In a standard $YDSE$ setup,a small transparent slab of thickness $t$ and refractive index $\mu = 1.5$ is placed along the path $AS_2$ (as shown in the figure). Given that the slab thickness $t = d/4$,where $d$ is the slit separation,and the distance from the source $A$ to the slits is not explicitly needed for the shift calculation,find the position of the central maxima on the screen relative to $O$. Assume the distance between the slits and the screen is $D$.

$A$ parallel beam of light of wavelength $500 \ nm$ is incident at an angle $30^o$ with the normal to the slit plane in a Young's double-slit experiment. The intensity due to each slit is $I_o$. Point $O$ is equidistant from $S_1$ and $S_2$. The distance between the slits is $1 \ mm$.

In a double slit arrangement,fringes are produced using light of wavelength $4800 \ \mathring A$. One slit is covered by a thin plate of glass of refractive index $1.4$ and the other with another glass plate of the same thickness but of refractive index $1.7$. By doing so,the central bright fringe shifts to the original fifth bright fringe from the center. The thickness of the glass plate is ...... $\mu m$.

In a double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern

In the figure shown,if a parallel beam of white light is incident on the plane of the slits,then the distance of the white spot on the screen from $O$ is [Assume $d << D, \lambda << d$].