In Young's double-slit experiment,the distance between the two slits is $3 \, cm$,the distance from the slits to the screen is $7 \, cm$,and the wavelength of light used is $1000 \, \mathring{A}$. Calculate the fringe width.

In Young's double slit experiment,the distance between the slits is $1 \,mm$ and the distance between the slit and the screen is $1 \,m$. If the $10^{th}$ fringe is $5 \,mm$ away from the central bright fringe,then the wavelength of the light used will be.....$\mathring A$.

In Young's double slit experiment,the wavelength of red light is $7800\,\mathring{A}$ and that of blue light is $5200\,\mathring{A}$. The value of $n$ for which the $n^{th}$ bright band due to red light coincides with the $(n + 1)^{th}$ bright band due to blue light is:

In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is:

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 $\text{mm}$)

Two slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)