In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

In Young's double-slit experiment,the width of one slit is twice that of the other. What is the ratio of the maximum to the minimum intensity?

In a Young's experiment,two coherent sources are placed $0.90 \, mm$ apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of $1 \, mm$ from the central fringe,the wavelength of monochromatic light used would be

In Young's double-slit experiment,the intensity at a point is $1/4$ of the maximum intensity. The angular position of this point is:

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is $d$ and $D$ is the distance between the source and the screen. The possible wavelengths used are inversely proportional to

Two light waves of wavelengths $800 \, nm$ and $600 \, nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \, m$ away from the plane of the slits. If the two slits are separated by $0.35 \, mm$,then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be $............. \, mm$.