To make the central fringe appear at the centre $O,$ a mica sheet of refractive index $\mu = 1.5$ is introduced. Choose the correct statement:

On placing a thin film of mica of thickness $12 \times 10^{-5} \text{ cm}$ in the path of one of the interfering waves in Young's double slit experiment using monochromatic light,the fringe pattern shifts through a distance equal to the width of a bright fringe. If the wavelength used is $\lambda = 6 \times 10^{-5} \text{ cm}$,the refractive index of mica is:

If a transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{-5} \, m$ is inserted in front of one of the slits of Young's Double Slit experiment,how much will be the shift in the interference pattern? The distance between the slits is $0.5 \, mm$ and that between slits and screen is $100 \, cm$. (Answer in $cm$)

Young's double slit experiment is carried out with two thin sheets of thickness $t = 10.4 \, \mu m$ each and refractive indices $\mu_1 = 1.52$ and $\mu_2 = 1.40$ covering the slits $S_1$ and $S_2$,respectively. If white light of range $400 \, nm$ to $780 \, nm$ is used,then which wavelength will form maxima exactly at point $O$,the centre of the screen?

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 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 $D$.