In Young's double-slit experiment, the fringe width is $\beta$. If the entire arrangement is placed in a liquid of refractive index $n$, what does the fringe width become?

In an interference experiment,the spacing between successive maxima or minima is (where the symbols have their usual meanings):

In a Young's double slit experiment,the intensities at two points,for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits,then $\frac{I_1 + I_2}{I_0} = \dots$

In Young's double-slit experiment,for a distance of $1.8 \lambda$ between the two slits,what is the maximum number of possible interference maxima? Here,$\lambda$ is the wavelength of light.

In Young's double-slit experiment,the fringe width is $12 \ mm$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$,then the fringe width becomes (in $mm$):

If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)