In a Young's double slit experiment,slits are separated by $0.5 \ mm$,and the screen is placed $150 \ cm$ away. $A$ beam of light consisting of two wavelengths,$650 \ nm$ and $520 \ nm$,is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is...... $mm$.

In Young's double-slit experiment,one of the slits is painted such that its intensity is half that of the other slit. Then:

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 a double-slit experiment,the distance between the slits is $d$. The screen is at a distance $D$ from the slits. If a bright fringe is formed opposite to one of the slits,its order $n$ is:

In a Young's double-slit experiment,sodium light of wavelength $5890 \ \mathring A$ is used,and the angular width of the fringe is found to be $0.20^\circ$. If the angular width is to be increased by $10\%$,what is the required change in the wavelength?

Two sources of light are $0.6 \, mm$ apart and the screen is placed at a distance of $1.2 \, m$ from them. $A$ light of wavelength $6000 \, \mathring{A}$ is used. The phase difference between the two light waves interfering on the screen at a point at a distance of $3 \, mm$ from the central bright band is: