Two slits are made one millimetre apart and the screen is placed one metre away from the slits. The fringe width when light of wavelength $500 \,nm$ is used is

Two slits separated by $0.5 \,mm$ are illuminated by light of wavelength $500 \,nm$. The screen is at a distance of $120 \,cm$ from the slits. The phase difference between the interfering waves at a point $3 \,mm$ on the screen from the central bright fringe is ........... .

In a Young's double slit experiment,if there is no initial phase difference between the light from the two slits,what is the path difference at a point on the screen corresponding to the fifth minimum?

Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ and $5^{\text{th}}$ bright band. The wavelength of light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:

In Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?

In a double slit experiment with monochromatic light,fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5 \times 10^{-2} \,m$ towards the slits,the change in fringe width is $3 \times 10^{-3} \,cm$. If the distance between the slits is $1 \,mm$,then the wavelength of the light will be . . . . . . $nm$.