In Young's double-slit experiment,if the ratio of the widths of the two slits is $4:9$,then the ratio of the maximum to minimum intensity will be:

In a $YDSE$ setup,the intensity of two coherent beams differs by $1\%$. If one of the beams has intensity $I$,then the intensity of the minima is:

In the Young's double slit experiment using a monochromatic light of wavelength $\lambda$,the path difference (in terms of an integer $n$) corresponding to any point having half the peak intensity is:

The amplitude of the light waves emerging from the two slits in Young's experiment is in the ratio of $2 : 3$. The ratio of the intensity of the minimum to that of the consecutive maximum will be:

If the intensity of each wave in the observed interference pattern in Young's double slit experiment is $I_0$,then for some point $P$ where the phase difference is $\phi$,the resultant intensity $I$ will be:

Consider a two-slit interference arrangement (see figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.