Two slits are made $10 \ mm$ apart and the screen is placed $1.5 \ m$ away. What is the fringe separation when a wavelength of $7000 \ \mathring{A}$ is used (in $\mu m$)?

If the sodium light in Young's double slit experiment is replaced by red light, the fringe width will

In Young's double slit experiment, the fringe width is $0.4 \text{ mm}$. What is the distance between $4^{th}$ dark band and $6^{th}$ bright band on the same side of the interference pattern (in $\text{ mm}$)?

In the figure, Young's double-slit experiment is shown. $Q$ is the position of the first bright fringe on the right side of $O$. $P$ is the $11^{th}$ fringe on the other side, as measured from $Q$. If the wavelength of the light used is $6000 \times 10^{-10} \text{ m}$, then $S_1B$ will be equal to:

In Young's double-slit experiment,the angular width of the fringes formed on the screen is $\pi / 200$. If light of wavelength $4800 \, \mathring{A}$ is used,find the distance between the slits.

Two wavelengths of light $\lambda_1$ and $\lambda_2$ are sent through a Young's double-slit experiment simultaneously. If the third-order bright fringe of $\lambda_1$ coincides with the fourth-order bright fringe of $\lambda_2$,then