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

In a $YDSE$,bi-chromatic light of wavelengths $400 \, nm$ and $560 \, nm$ are used. The distance between the slits is $0.1 \, mm$ and the distance between the plane of the slits and the screen is $1 \, m$. The minimum distance between two successive regions of complete darkness is......$mm$.

In Young's double slit experiment,the distance between the two slits is $0.1 \,mm$,the distance between the slits and the screen is $1 \,m$ and the wavelength of the light used is $600 \,nm$. The intensity at a point on the screen is $75 \%$ of the maximum intensity. Find the smallest distance in $mm$ of this point from the central fringe.

In Young's double slit experiment, the width of the second slit is double the width of the first slit. Consequently, the amplitude of the light from the two slits is different. If $I_m$ is the maximum intensity, the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is given by:

In Young's double slit experiment,the angular width of fringes is $0.20^o$ for sodium light of wavelength $5890 \ \mathring A$. If the complete system is dipped in water,then the angular width of the fringes becomes....$^o$

In an experiment, light passing through two slits separated by a distance of $0.3 \,mm$ is projected onto a screen placed at $1 \,m$ from the plane of the slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 \,mm$. The wavelength of light in $nm$ is