Angular width $(\beta)$ of the central maximum of a diffraction pattern on a single slit does not depend upon

$A$ light source with a wavelength of $5000 \, \mathring A$ produces a single-slit diffraction pattern. The first minimum in the diffraction pattern is observed at a distance of $5 \, mm$ from the central maximum. The distance between the slit and the screen is $2 \, m$. Find the width of the slit. (in $, mm$)

$A$ beam of light of $\lambda = 600 \, nm$ from a distant source falls on a single slit $1 \, mm$ wide and the resulting diffraction pattern is observed on a screen $2 \, m$ away. The distance between the first dark fringes on either side of the central bright fringe is:

Define diffraction.

$A$ light wave of wavelength $\lambda$ is incident on a slit of width $d$. The resulting diffraction pattern is observed on a screen at a distance $D$. If the linear width of the principal maxima is equal to the width of the slit,then the distance $D$ is:

$A$ plane wavefront of wavelength $6 \times 10^{-7} \,m$ is incident on a slit of width $0.4 \,mm$. $A$ convex lens of focal length $0.8 \,m$ is placed behind the slit to form a diffraction pattern on a screen. What is the linear width of the second secondary maximum in $mm$?