$A$ 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:

If a narrow slit of width $2 \, mm$ is illuminated by monochromatic light of wavelength $500 \, nm$,then the distance between the first minima on both sides on a screen at a distance of $1 \, m$ is ....... $mm$.

$A$ parallel beam of light of wavelength $\lambda$ is incident normally on a narrow slit. $A$ diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern,the phase difference between the rays coming from the two edges of the slit is:

$A$ slit of width $a$ is illuminated by red light of wavelength $6500 \, \mathring{A}$. The first minimum will fall at $\theta = 30^{\circ}$ if $a$ is equal to:

Light of wavelength $5000 \, \mathring{A}$ is incident on a single slit such that the first minima is formed at a distance $5 \, mm$ from the centre. If the screen is placed $2 \, m$ away,then find the width of the slit in $mm$.

In a Fraunhofer diffraction at a single slit of width $d$ and incident light of wavelength $5500 \text{ Å}$,the first minimum is observed at an angle $30^{\circ}$. The first secondary maxima are observed at an angle $\theta=$