Light of wavelength $6000 \ \mathring{A}$ is incident on a slit. The width of the slit is $0.30 \ mm$. A screen is placed at a distance of $2 \ m$ from the slit. Find the width of the central maximum in $mm$.

If $n$ represents the order of a half-period zone,the area of this zone is approximately proportional to $n^m$,where $m$ is equal to:

In a single slit diffraction pattern,the distance between the plane of the slit and the screen is $1.3 \ m$. The width of the slit is $0.65 \ mm$ and the second maximum is formed at a distance of $2.6 \ mm$ from the center of the screen. The wavelength of light used is: (in $Å$)

Light of wavelength $5000 \text{ Å}$ is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of $5 \text{ mm}$ from the central maximum on a screen placed at a distance of $2 \text{ m}$ from the slit. The width of the slit is: (in $\text{ cm}$)

$A$ slit of width $d$ is illuminated by light of wavelength $\lambda$. The angular width of the central maximum is:

In the experiment of diffraction at a single slit,if the slit width is decreased,the width of the central maximum