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:

In a Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $K$. What will be the intensity at a point where the path difference is $\lambda / 4$?

In a Young's double slit experiment, the slits are separated by $0.2 \text{ mm}$ and the screen is placed $2.0 \text{ m}$ away. The distance between the central bright fringe and the third bright fringe is measured to be $1.5 \text{ cm}$. Determine the wavelength of light used in the experiment.

In the Young's double slit experiment with sodium light,the slits are $0.589 \ m$ apart. The angular separation of the third maximum from the central maximum will be (given $\lambda = 589 \ nm$):

In Young's double-slit experiment,when two light waves form the third minimum,they have:

In a Young's double-slit experiment,$D$ is the distance of the screen from the slits and $d$ is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is the same as that due to a single slit is equal to: