Two parallel slits $0.6 \, mm$ apart are illuminated by a light source of wavelength $6000 \, \mathring{A}$. The distance between two consecutive dark fringes on a screen $1 \, m$ away from the slits is ........ $mm$.

In a double-slit experiment using light of wavelength $600 \; nm$,the angular width of a fringe formed on a distant screen is $0.1^{\circ}$. What is the spacing between the two slits?

In a double-slit interference experiment, the fringe width obtained with light of wavelength $5900 \ \mathring{A}$ was $1.2 \ \text{mm}$ for parallel narrow slits placed $2 \ \text{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-a-half times the previous value, then the fringe width is: (in $\text{mm}$)

The distance between two slits is $3 \ mm$ and the screen is placed at $2 \ m$ distance. When blue-green light of wavelength $500 \ nm$ is used, then the distance between two consecutive fringes will be (in $mm$)?

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

$A$ source simultaneously emitting light at two wavelengths $400 \,nm$ and $800 \,nm$ is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is $I_{0}$,then the maximum intensity that can be observed at any point on the screen is