In Young's double slit experiment,carried out with light of wavelength $5000 \ \mathring{A}$,the distance between the slits is $0.3 \ \text{mm}$ and the screen is at $200 \ \text{cm}$ from the slits. The central maximum is at $x=0 \ \text{cm}$. The value of $x$ for the third maxima is ............. $\text{mm}$.

In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

In Young's double-slit experiment,the position of the $5^{\text{th}}$ bright fringe from the central maximum is $5\,cm$. The distance between the slits and the screen is $1\,m$ and the wavelength of the monochromatic light used is $600\,nm$. The separation between the slits is $........\mu m$.

Two slits are made one millimetre apart and the screen is placed one metre away. What is the fringe separation (in $mm$) when blue-green light of wavelength $500 \; nm$ is used?

In Young's double slit experiment,the intensity on the screen at a point where path difference is $\lambda$ is $K.$ What will be the intensity at the point where path difference is $\lambda /4$?

Using Young's double slit experiment,a monochromatic light of wavelength $5000 \,\mathring A$ produces fringes of fringe width $0.5 \,mm$. If another monochromatic light of wavelength $6000 \,\mathring A$ is used and the separation between the slits is doubled,then the new fringe width will be ............... $mm$.