What happens to the fringe width in Young's double-slit experiment if it is performed in glycerine instead of air?

Light consisting of wavelengths $6000\,\mathring{A}$ and $5500\,\mathring{A}$ falls on the double slits in $YDSE$. The $n^{th}$ order bright fringe of $\lambda_1 = 6000\,\mathring{A}$ is found to coincide with the $m^{th}$ order bright fringe of $\lambda_2 = 5500\,\mathring{A}$. The smallest values of $n$ and $m$ are respectively:

In Young's double-slit experiment,if $L$ is the distance between the slits and the screen upon which the interference pattern is observed,$x$ is the average distance between adjacent fringes,and $d$ is the slit separation,then the wavelength of light is given by:

In a Young's double-slit experiment, the distance between two coherent sources is $0.90 \, mm$ and the distance from the sources to the screen is $1 \, m$. If the distance of the second bright fringe from the center of the central bright fringe is $1 \, mm$, find the wavelength of the light used.

In Young's experiment,the distance between slits is $0.28 \,mm$ and the distance between the slits and the screen is $1.4 \,m$. The distance between the central bright fringe and the third bright fringe is $0.9 \,cm$. What is the wavelength of the light used in $\mathring{A}$?

Young's double-slit experiment is performed using green light,red light,and blue light,one at a time. The fringe widths recorded are $\beta_G, \beta_R,$ and $\beta_B$ respectively. Then: