If the sodium light in Young's double slit experiment is replaced by red light, the fringe width will

$A$ 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?

The width of the fringe is $2\,mm$ on the screen in a double-slit experiment for light of wavelength $400\,nm$. The width of the fringe for light of wavelength $600\,nm$ will be $..............\,mm$.

In Young's double slit experiment, for the $n^{th}$ dark fringe $(n = 1, 2, 3, \dots)$ the phase difference of the interfering waves in radian will be

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 an interference experiment,the $n^{\text{th}}$ bright fringe for light of wavelength $\lambda_1$ $(n=0, 1, 2, 3, \ldots)$ coincides with the $m^{\text{th}}$ dark fringe for light of wavelength $\lambda_2$ $(m=1, 2, 3, \ldots)$. The ratio $\frac{\lambda_1}{\lambda_2}$ is