In a Young's double slit experiment illuminated by monochromatic light,the intensity reaching the screen from each slit is $I$. What is the resultant intensity at a point where the phase difference is $60^{\circ}$?

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.

In $YDSE$,$16$ fringes occupy a certain region of space with light of wavelength $6000\,\mathring{A}$. If $24$ fringes occupy the same region,the wavelength of the new light is........$\mathring{A}$.

Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate 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 . . . . . . $mm$.