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:

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^{\circ}$ in a single slit diffraction pattern recorded using $628 \ nm$ light,then the width of the slit is . . . . . . $\mu m$.

When the distance between the slit and the screen is doubled,what happens to the angular separation between the fringes in a slit diffraction experiment?

Light of wavelength $6328 \, \mathring{A}$ is incident on a slit of width $0.2 \, mm$. Find the angular width of the central maximum on a screen at a distance of $9 \, m$ in degrees.

If $\lambda = 6000 \, \mathring{A}$ and $a = 18 \times 10^{-5} \, \text{cm}$,find the angular width of the central maximum in degrees $(^\circ)$.