In Young's double slit experiment, the fringe width is $0.4 \text{ mm}$. What is the distance between $4^{th}$ dark band and $6^{th}$ bright band on the same side of the interference pattern (in $\text{ mm}$)?

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width is

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\theta$ is the angle $S_1CS_2$ and $\lambda$ is the wavelength,the fringe width will be

In $Y.D.S.E.$ using red and blue lights of wavelengths $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$,the $n^{th}$ red fringe coincides with the $(n + 1)^{th}$ blue fringe. The value of $n$ is:

In Young's double-slit experiment,when light of wavelength $\lambda_1 = 5898 \ \mathring{A}$ is used,$92$ fringes are observed in a certain region. If light of wavelength $\lambda_2 = 5461 \ \mathring{A}$ is used instead,how many fringes will be observed in the same region?

$A$ Young's double slit experiment apparatus has slits separated by $0.2 \ mm$ and a screen $60 \ cm$ away from the slits. The whole apparatus is immersed in a liquid medium of refractive index $\mu = \frac{11}{9}$ and the slits are illuminated with green light $(\lambda = 550 \ nm$ in vacuum$)$. Find the fringe width of the pattern formed on the screen. (in $mm$)