Young's experiment demonstrates that . . . . . .

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 a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring A$ and $5200 \, \mathring A$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. Find the distance of the third bright fringe from the central maximum for the wavelength $6500 \, \mathring A$ in $mm$.

In Young's double-slit experiment,the intensity at a point is $(1/4)$ of the maximum intensity. The angular position of this point is:

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

In a Young's double slit experiment,the distance between the two identical slits is $6.1$ times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is