In a $Young's$ double-slit experiment,light of wavelength $6000 \, \mathring{A}$ is used. If the third dark fringe is formed at point $P$ on the screen,then the path difference $(S_1P - S_2P)$ is ......... $\mu m$.

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

The graph shows the variation of fringe width $(X)$ versus the distance of the screen from the plane of the slits $(D)$ in Young's double-slit experiment (keeping other parameters constant,where $d$ is the distance between the slits). The wavelength of light used can be calculated as:

In Young's double-slit experiment,the distance between the two slits is $2 \times 10^{-3} \, m$ and the distance between the slits and the screen is $2.5 \, m$. The wavelength of the light used ranges from $2000 \, \mathring{A}$ to $9000 \, \mathring{A}$. What wavelength (in $\mathring{A}$) will form a bright fringe at a distance of $10^{-3} \, m$ from the central maximum?

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 $\text{mm}$)

In the Young's double slit experiment,the ratio of intensities of bright and dark fringes is $9$. This means that