In the experiment of diffraction due to a single slit,if the slit width is decreased,the width of the central maximum

At the first minimum adjacent to the central maximum of a single-slit diffraction pattern,the phase difference between the Huygens' wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

In a single slit diffraction pattern,the distance between the plane of the slit and the screen is $1.3 \ m$. The width of the slit is $0.65 \ mm$ and the second maximum is formed at a distance of $2.6 \ mm$ from the center of the screen. The wavelength of light used is: (in $Å$)

$A$ circular disc is placed in front of a narrow source. When the point of observation is at a distance of $1 \, m$ from the disc,the disc covers the first $HPZ$ (Half Period Zone). The intensity at this point is $I_0$. The intensity at a point at a distance of $25 \, cm$ from the disc will be:

In Fraunhofer diffraction by a single slit,the width of the slit is $0.01 \ cm$. If the wavelength of the light incident normally on the slit is $5000 \ \mathring{A}$,the angular distance of the second maxima from the midline of the central maxima is . . . . . . $\text{rad}$.

$A$ beam of light having wavelength $5400 \text{ Å}$ from a distant source falls on a single slit $0.96 \text{ mm}$ wide and the resultant diffraction pattern is observed on a screen $2 \text{ m}$ away. What is the distance between the first dark fringe on either side of the central bright fringe (in $\text{mm}$)?