Interference was observed in an interference chamber when air was present. Now,the chamber is evacuated and if the same light is used,a careful observer will see:

In Young's double-slit experiment,one of the slits is painted such that its intensity is half that of the other slit. Then:

In a Young's double slit experiment,$12$ fringes are observed to be formed in a certain segment of the screen when light of wavelength $600 \ nm$ is used. If the wavelength of light is changed to $400 \ nm$,the number of fringes observed in the same segment of the screen is:

In Young's experiment,the distance between two slits is $d/3$ and the distance between the screen and the slits is $3D$. The number of fringes in $1/3 \ m$ on the screen,formed by monochromatic light of wavelength $3\lambda$,will be

In Young's double slit experiment,the slits are $0.5 \, mm$ apart and the interference is observed on a screen at a distance of $100 \, cm$ from the slits. It is found that the $9^{th}$ bright fringe is at a distance of $7.5 \, mm$ from the $2^{nd}$ dark fringe on the same side of the central fringe. The wavelength of the light used is ..... $\mathring{A}$.

In Young's double-slit interference experiment,the ratio of the slit widths is $1:25$. Find the ratio of the maximum to minimum intensity in the interference pattern.