Light of wavelength $589.3 \, nm$ is incident normally on a slit of width $0.1 \, mm$. What will be the angular width of the central diffraction maximum in degrees?

The angular deviation of the $5^{th}$ order dark fringe is $12^{\circ}$ in a single slit experiment. If the width of the slit is $9 \mu m$, then the wavelength of the incident light is: (in $Å$)

"The person standing behind the open door inside the room can hear the voice of the person standing on the other side of the door but they cannot see each other". Give an experiment based on the diffraction for this statement.

Determine the angular separation between the central maximum and the first-order maxima of the diffraction pattern due to a single slit of width $0.25\, mm$ when light of wavelength $5890\,\mathring{A}$ is incident on it.

$A$ microwave of wavelength $2.0 \ cm$ falls normally on a slit of width $4.0 \ cm$. The angular spread of the central maxima of the diffraction pattern obtained on a screen $1.5 \ m$ away from the slit,will be: (in $^{\circ}$)

In the diffraction pattern, the first maximum is at $30^{\circ}$, when a monochromatic light of wavelength $\lambda$ is incident on a slit of width $a$. For the same wavelength, if the slit width is changed, so that the first maximum is at $45^{\circ}$, by how much is the slit width changed?