In a Young's double-slit experiment,the distance between the two slits is twice the wavelength of the light used. Find the total number of bright fringes formed on the screen.

$A$ beam of light consisting of two wavelengths $7000 \; \mathring{A}$ and $5500 \; \mathring{A}$ is used to obtain an interference pattern in Young's double-slit experiment. The distance between the slits is $2.5 \; mm$ and the distance between the plane of the slits and the screen is $150 \; cm$. The least distance from the central fringe,where the bright fringes due to both the wavelengths coincide,is $n \times 10^{-5} \; m$. The value of $n$ is .............

What happens to the fringe width in Young's double-slit experiment when it is performed in water instead of air?

In an interference experiment, the third bright fringe is obtained at a point on the screen with light of wavelength $700 \, nm$. What should be the wavelength of the light source in order to obtain the $5^{th}$ bright fringe at the same point (in $nm$)?

When a thin metal plate is placed in the path of one of the interfering beams of light in a Young's double-slit experiment,what happens to the interference pattern?

Two slits are separated by a distance of $0.5\, mm$ and illuminated with light of $\lambda = 6000\ \mathring A$. If the screen is placed $2.5\, m$ from the slits,the distance of the third bright fringe from the centre will be........$mm$.