In Young's double slit interference experiment,the slit widths are in the ratio $1 : 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is

Two slits are separated by a distance of $0.5\, mm$ and illuminated with light of $\lambda = 6000\ \mathring A$. If the screen is placed $2.5\, m$ from the slits,the distance of the third bright fringe from the centre will be........$mm$.

In a $YDSE$ experiment,if a slab whose refractive index can be varied is placed in front of one of the slits,then the variation of resultant intensity at the mid-point of the screen with $\mu$ will be best represented by $(\mu \geq 1)$. [Assume slits of equal width and there is no absorption by the slab]

Assertion: $A$ white source of light during interference forms only white and black fringes.
Reason: Width of fringe is inversely proportional to the wavelength of the light used.

In two different Young's double-slit experiments,the fringe width is the same when the ratio of wavelengths is $1:2$. If the ratio of the distance between the slits in the two cases is $2:1$,then the ratio of the distance between the slits and the screen in the two experiments is:

Write down the formula for the distance between two consecutive bright fringes or two consecutive dark fringes in Young's Double Slit Experiment.