In a single-slit diffraction pattern,if the source of light is replaced by a source of shorter wavelength,the width of the central maximum will:

To observe diffraction,the size of the obstacle

$A$ single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $\lambda$. If the second and fourth minima in the diffraction pattern at a distance $1\,m$ from the slit are at $3\,cm$ and $6\,cm$ respectively from the central maximum,what is the width of the central maximum in $cm$ (i.e.,distance between the first minimum on either side of the central maximum)?

$A$ diffraction pattern is obtained by using a beam of red light. If the red light is replaced by blue light,then:

$A$ 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 $\text{mm}$)?

When the frequency of the light used is changed from $4 \times 10^{14} \ s^{-1}$ to $5 \times 10^{14} \ s^{-1}$, the angular width of the principal (central) maximum in a single slit Fraunhofer diffraction pattern changes by $0.6 \ \text{radian}$. What is the width of the slit? (Assume that the experiment is performed in vacuum.)