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 $mm$.

$A$ 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:

The waves of $600 \ \mu m$ wavelength are incident normally on a slit of $1.2 \ mm$ width. The value of diffraction angle corresponding to the first minima will be (in radian) -

In the diffraction pattern of a single slit,the width of the secondary maxima compared to the central maximum is:

The condition for obtaining secondary maxima in the diffraction pattern due to a single slit is:

$A$ beam of light of $\lambda = 600 \, nm$ from a distant source falls on a single slit $1 \, mm$ wide and the resulting diffraction pattern is observed on a screen $2 \, m$ away. The distance between the first dark fringes on either side of the central bright fringe is: