In Young’s double slit experiment,a minimum is obtained when the phase difference of superimposing waves is:

In Young's double slit experiment,the light emitted from the source has $\lambda = 6.5 \times 10^{-7} \, m$ and the distance between the two slits is $1 \, mm$. The distance between the screen and the slits is $1 \, m$. The distance between the third dark fringe and the fifth bright fringe will be ......... $mm$.

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.

$A$ Young's double-slit experiment is performed using monochromatic light of wavelength $\lambda$. The intensity of light at a point on the screen,where the path difference is $\lambda$,is $K$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ is given by $\frac{nK}{12}$,where $n$ is an integer. The value of $n$ is $......$

In Young's double-slit experiment,if the amplitudes of the interfering waves are not equal,then . . . . .

In a Young's double slit experiment,the angular width of a fringe formed on a distant screen is $0.1 \ radian$. Find the distance between the two slits in $\mu m$,if the wavelength of light used is $6000 \ \mathring{A}$.