In a Young's double-slit experiment,the distance between the two slits is $d = \lambda / 4$,where $\lambda$ is the wavelength of the light used. The initial phase difference is $\pi / 4$. What is the intensity at $\theta = 30^o$?

In an interference experiment,the spacing between successive maxima or minima is

In Young's double slit experiment,the distance between the slits is reduced to half and the distance between the slit and the screen is doubled,then the fringe width:

In Young's double slit experiment,the ratio of intensity of maxima and minima in the interference experiment is $25 : 9$. The ratio of the widths of the two slits is:

In a Young's double slit experiment, a small detector measures an intensity of illumination of $I$ units at the centre of the fringe pattern. If one of the two (identical) slits is now covered, the measured intensity will be

Interference fringes are observed on a screen by illuminating two thin slits $1 \, mm$ apart with a light source $(\lambda = 632.8 \, nm)$. The distance between the screen and the slits is $100 \, cm$. If a bright fringe is observed on a screen at a distance of $1.27 \, mm$ from the central bright fringe,then the path difference between the waves,which are reaching this point from the slits,is close to $.... \mu m$.