The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

Two wavelengths $\lambda_1 = 450 \ nm$ and $\lambda_2 = 650 \ nm$ are used in Young's double slit experiment. The minimum order of fringe produced by $\lambda_2$ which overlaps with a fringe produced by $\lambda_1$ is $n$. The value of $n$ is . . . . . . .

In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

In Young's double slit experiment,if monochromatic light is replaced by white light:

$A$ mixture of light,consisting of wavelengths $590 \ nm$ and an unknown wavelength,illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further,it is observed that the $3^{rd}$ bright fringe of the known light coincides with the $4^{th}$ bright fringe of the unknown light. From this data,the wavelength of the unknown light is ...... $nm$.