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

$A$ screen is placed $0.5 \,m$ away from a single slit which is illuminated by a monochromatic light of wavelength $6000 \text{ Å}$. If the distance between the first and third minima in the diffraction pattern on the screen is $3 \,mm$, then the slit width is: (in $\,mm$)

The phenomenon of diffraction of light was discovered by

In a single slit diffraction of light of wavelength $\lambda$ by a slit of width $e$,the size of the central maximum on a screen at a distance $b$ is

Light of wavelength $6000 \, \mathring A$ is incident on a slit of width $0.30 \, mm$. $A$ screen is placed at a distance of $2 \, m$ from the slit. Find the position of the first minimum.

$A$ single slit diffraction pattern is formed with light of wavelength $6384 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength $\lambda_0$. The value of $\lambda_0$ is (in $Å$)