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 $D=$

The position of the direct image obtained at $O$,when a monochromatic beam of light is passed through a plane transmission grating at normal incidence,is shown in the figure. The diffracted images $A, B$,and $C$ correspond to the first,second,and third-order diffraction. When the source is replaced by another source of shorter wavelength,then:

$A$ single slit diffraction pattern is formed with white light. For what wavelength of light, the $3^{\text{rd}}$ secondary maximum in the diffraction pattern coincides with the $2^{\text{nd}}$ secondary maximum in the pattern of red light of wavelength $6000 \text{ Å}$ (in $\text{ Å}$)?

If $I_0$ is the intensity of the principal maximum in the single slit diffraction pattern,then what will be the intensity when the slit width is doubled?

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

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