In $YDSE$ with two identical slits,when the upper slit is covered with a transparent sheet of mica,the second-order minima is observed above the center of the screen and the second-order maxima below it. Which of the following cannot be a possible value of the phase difference caused by the mica sheet?

Two slits,$4 \, mm$ apart,are illuminated by light of wavelength $6000 \, \mathring{A}$. What will be the fringe width on a screen placed $2 \, m$ from the slits in $mm$?

This question has Statement-$1$ and Statement-$2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement-$1$: In Young's double slit experiment,the number of fringes observed in the field of view is small with longer wavelength of light and is large with shorter wavelength of light.
Statement-$2$: In the double slit experiment,the fringe width depends directly on the wavelength of light.

In a Young's double slit experiment,a laser light of $560\,nm$ produces an interference pattern with consecutive bright fringes' separation of $7.2\,mm$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1\,mm$. The wavelength of the second light is $......nm$.

In a Young's double slit experiment,the path difference,at a certain point on the screen,between two interfering waves is $1/8^{th}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

The figure shows a double slit experiment where $P$ and $Q$ are the slits. The path lengths $PX$ and $QX$ are $n\lambda$ and $(n + 2)\lambda$ respectively,where $n$ is a whole number and $\lambda$ is the wavelength. Taking the central fringe as zero,what is formed at $X$?