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:

The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

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:

In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

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,if the distance between the slits is twice the wavelength,the number of possible interference maxima is: