In Young's double-slit experiment,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{87}{2}\right) \lambda$. The band obtained at that point is

Light of wavelength $6000 \,\mathring A$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. What is the distance between two consecutive minima in $mm$?

The maximum intensity in Young's double slit experiment is $I_0$. The distance between the slits is $d = 5\lambda$,where $\lambda$ is the wavelength of the monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D = 10d$?

In Young's double-slit experiment,when two light waves form the third minimum,they have:

At two points $P$ and $Q$ on the screen in Young's double-slit experiment,waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda}{4}$ respectively. The ratio of intensities at $P$ and $Q$ will be