If the slit widths of $YDSE$ are in the ratio $1:4$,then the ratio of maximum to minimum intensity on the screen is: (in $:1$)

If the Young's double slit experiment is performed with white light,then which of the following is not true?

In a Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $K$. What will be the intensity at a point where the path difference is $\lambda / 4$?

In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

Four monochromatic and coherent sources of light,emitting waves in phase of wavelength $\lambda$,are placed at the points $x = 0, d, 2d$,and $3d$ on the $x$-axis. Then:

In Young's double-slit experiment,the position of the $5^{\text{th}}$ bright fringe from the central maximum is $5\,cm$. The distance between the slits and the screen is $1\,m$ and the wavelength of the monochromatic light used is $600\,nm$. The separation between the slits is $........\mu m$.