In Young's double-slit experiment,when two light waves form the third minimum,the path difference between them is:

In Young's double slit experiment,white light is used. The separation between the slits is $b$. The screen is at a distance $d$ $(d >> b)$ from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

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

$A$ beam of light consisting of two wavelengths $6300 \, \mathring{A}$ and $\lambda \, \mathring{A}$ is used to obtain interference fringes in a Young's double-slit experiment. If the $4^{th}$ bright fringe of $6300 \, \mathring{A}$ coincides with the $5^{th}$ dark fringe of $\lambda \, \mathring{A}$,the value of $\lambda$ (in $\mathring{A}$) is:

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,$I/I_0$ is equal to