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 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 a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

$A$ double slit interference experiment performed with a light of wavelength $600 \ nm$ forms an interference fringe pattern on a screen with $10^{\text{th}}$ bright fringe having its centre at a distance of $10 \ mm$ from the central maximum. The distance of the centre of the same $10^{\text{th}}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength $660 \ nm$ would be . . . . . . $mm$.

In a Young's double-slit experiment,interference fringes are obtained on a screen placed at a certain distance from the slits using monochromatic light. If the screen is moved towards the slits by $5 \times 10^{-2} \, m$,the fringe width changes by $3 \times 10^{-5} \, m$. If the distance between the slits is $10^{-3} \, m$,find the wavelength of the light.

Young's double-slit experiment is performed using microwaves of wavelength $\lambda = 3 \, cm$. The distance between the plane of the slits and the screen is $D = 100 \, cm$,and the distance between the slits is $d = 5 \, cm$. Find $(a)$ the number of maxima and $(b)$ their positions on the screen.