In a Young's double-slit experiment, interference fringes are obtained on a screen at a distance of $1 \, m$ using light of wavelength $6000 \, \mathring{A}$. The distance between the slits is $1 \, mm$. The fringe width is:

In Young's double slit experiment,the phase difference between the light waves reaching the third bright fringe from the central fringe is: $(\lambda = 6000 \ \mathring{A})$

In Young's double slit experiment,the slits are $2\, mm$ apart and are illuminated by light of two wavelengths $\lambda_1 = 12000\, \mathring{A}$ and $\lambda_2 = 10000\, \mathring{A}$. At what minimum distance from the common central bright fringe on the screen $2\, m$ from the slits will a bright fringe from one interference pattern coincide with a bright fringe from the other? (in $mm$)

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.

In Young's double slit experiment,the fringes are displaced by a distance $x$ when a glass plate of refractive index $1.5$ is introduced in the path of one of the beams. When this plate is replaced by another plate of the same thickness,the shift of fringes is $(3/2)x$. The refractive index of the second plate is

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