If the separation between slits in Young's double slit experiment is reduced to $\frac{1}{3}rd,$ the fringe width becomes $n$ times. The value of $n$ is

The graph shows the variation of fringe width $(X)$ versus the distance of the screen from the plane of the slits $(D)$ in Young's double-slit experiment (keeping other parameters constant,where $d$ is the distance between the slits). The wavelength of light used can be calculated as:

In a Young's double slit experiment, the slits are $2 \,mm$ apart and are illuminated with a mixture of two wavelengths ${\lambda _1} = 750 \,nm$ and ${\lambda _2} = 900 \,nm$. The minimum distance from the common central bright fringe on a screen $2 \,m$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is.....$mm$

Two wavelengths $\lambda_1 = 450 \ nm$ and $\lambda_2 = 650 \ nm$ are used in Young's double slit experiment. The minimum order of fringe produced by $\lambda_2$ which overlaps with a fringe produced by $\lambda_1$ is $n$. The value of $n$ is . . . . . . .

In $Y.D.S.E$,the ratio of intensity of maxima and minima is $25 : 9$. The ratio of slit width is

In Young's double slit experiment,the slits are $0.5\, mm$ apart and interference pattern is observed on a screen placed at a distance of $1.0\, m$ from the plane containing the slits. If the wavelength of the incident light is $6000\ \mathring A$,then the separation between the third bright fringe and the central maxima is......$mm$.