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:

$A$ 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 ...... $nm$.

In a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

In Young's double slit experiment,the distance between two sources is $0.1 \, mm$. The distance of the screen from the sources is $20 \, cm$. The wavelength of light used is $5460 \, \mathring{A}$. The angular position of the first dark fringe is: (in $^\circ$)

In Young's double slit experiment,we get $60$ fringes in the field of view using monochromatic light of wavelength $4000 \; \mathring{A}$. If we use monochromatic light of wavelength $6000 \; \mathring{A}$,then the number of fringes obtained in the same field of view is

Two identical narrow slits $S_1$ and $S_2$ are illuminated by light of wavelength $\lambda$ from a point source $P$. If,as shown in the diagram,the light is then allowed to fall on a screen,and if $n$ is a positive integer,the condition for destructive interference at $Q$ is that