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

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is $5 \,mm$. The screen on which the diffraction pattern is obtained is at a distance of $80 \,cm$ from the slit. The wavelength used is $6000 \mathring{A}$. The width of the slit is (in $\,mm$)

Light of wavelength $\lambda$ is incident on a slit of width $a$. The resulting diffraction pattern is observed on a screen placed at a distance $D$. If the linear width of the central maximum is equal to the width of the slit,then $D = $ . . . . . . .

In a single slit diffraction experiment,the width of the slit is made double its original width. Then the central maximum of the diffraction pattern will become

$A$ 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:

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,