In $YDSE$,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. Then,the fringe width:

In a Young's double-slit experiment,two slits are separated by a distance of $3 \, mm$ and 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 fringe.

$A$ beam of light consisting of wavelengths $650 \ nm$ and $550 \ nm$ illuminates the Young's double slits with separation of $2 \ mm$ such that the interference fringes are formed on a screen, placed at a distance of $1.2 \ m$ from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is . . . . . . $\times 10^{-5} \ m$.

In a Young's double-slit experiment,sources of equal intensities are used. The distance between the slits is $d$ and the wavelength of light used is $\lambda$ (where $\lambda \ll d$). Find the angular separation of the nearest points on either side of the central maximum where the intensity becomes half of the maximum value.

In a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

If the source of light used in a Young's double slit experiment is changed from red to violet: