The ratio of the distance of the $n^{\text{th}}$ bright band and the $m^{\text{th}}$ dark band from the central bright band in an interference pattern is

In a Young's double-slit experiment, interference fringes are obtained on a screen at a distance of $1 \, m$ using light of wavelength $6000 \, \mathring{A}$. The distance between the slits is $1 \, mm$. The fringe width is:

In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

Consider a Young's double slit experiment as shown in the figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$?

In a $YDSE$ apparatus,if we use white light,then: