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 $(S_1P - S_2P)$

$A$ transparent film $(\mu=1.45)$ of thickness $0.02 \ mm$ is placed on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength $620 \ nm$. How many fringes will cross through the center if the film is removed?

The figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits,the central maximum shifts by a distance equal to $n$ fringe widths. If the wavelength of light used is $\lambda$,then $t$ will be:

In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

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 $D$.

Interference fringes were produced in Young's double slit experiment using light of wavelength $5000 \, Å$. When a film of material $2.5 \times 10^{-3} \, cm$ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to $20$ fringe widths. The refractive index of the material of the film is