In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

In Young's double-slit experiment,the separation between the two slits is $3 \, mm$ and they are illuminated by light of wavelength $480 \, nm$. The screen is at a distance of $2 \, m$ from the plane of the slits. Find the separation between the $8^{th}$ bright fringe and the $3^{rd}$ dark fringe with respect to the central bright fringe.

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:

In Young's double slit experiment,$62$ fringes are seen in the visible region for sodium light of wavelength $5893 \ \mathring{A}$. If violet light of wavelength $4358 \ \mathring{A}$ is used in place of sodium light,then the number of fringes seen will be:

$A$ 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 . . . . . . $mm$.

Two ideal slits $S_1$ and $S_2$ are at a distance $d$ apart and are illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_2$ as shown. The distance between the plane of the slits and the source slit is $D$. $A$ screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is