Light of wavelength $\lambda$ is incident on a single slit of width $a$ and the distance between the slit and the screen is $D$. In the diffraction pattern, if the slit width is equal to the width of the central maximum, then $D$ is equal to:

$A$ parallel monochromatic beam of light is incident normally on a narrow slit. $A$ diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first secondary maximum of the diffraction pattern,the phase difference between the rays coming from the edges of the slit is

$A$ single slit of width $a$ is illuminated by violet light of wavelength $400 \ nm$ and the width of the diffraction pattern is measured as $y.$ When half of the slit width is covered and illuminated by yellow light of wavelength $600 \ nm$,the width of the diffraction pattern is

In the far-field diffraction pattern of a single slit under polychromatic illumination,the first minimum with the wavelength ${\lambda _1}$ is found to be coincident with the third maximum at ${\lambda _2}$. So,

The first diffraction minima due to a single slit diffraction is at $\theta = 30^o$ for a light of wavelength $5000 \,\mathring{A}$. The width of the slit is

Two wavelengths of sodium light $590 \,nm$ and $596 \,nm$ are used one after another to study diffraction due to a single slit of aperture $2 \times 10^{-6} \,m$. The distance between the slit and the screen is $1.5 \,m$. The separation between the positions of the first maximum of the diffraction pattern obtained in the two cases is: (in $\,mm$)