The fringe width in an interference pattern is $X$. The distance between the sixth dark fringe from one side of the central bright band to the fourth bright fringe on the other side is: (in $X$)

In Young's double slit experiment,the seventh maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum,and the same maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $d_1/d_2$ is:

$A$ double slit experiment is immersed in water of refractive index $n = 1.33$. The slit separation is $d = 1 \text{ mm}$, and the distance between the slit and the screen is $D = 1.33 \text{ m}$. The slits are illuminated by light of wavelength $\lambda_0 = 6300 \text{ Å}$ in vacuum. The fringe width is:

In a Young's double-slit experiment,the slits are placed $0.320 \, mm$ apart. Light of wavelength $\lambda = 500 \, nm$ is incident on the slits. The total number of bright fringes that are observed in the angular range $-30^\circ \le \theta \le 30^\circ$ is:

For sustained interference fringes in a double-slit experiment,the essential condition$(s)$ is/are:
$1$. Sources must be coherent.
$2$. The intensities of the two sources must be equal.
Here,the correct option$(s)$ is/are:

In an interference experiment,let $D$ be the distance between the slit and the eyepiece. When the distance between two virtual sources is changed from $d_{A}$ to $d_{B}$,the fringe width changes from $Z_{A}$ to $Z_{B}$. The ratio $Z_{A} / Z_{B}$ is: