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 $mm$.

Light of wavelength $580 \text{ nm}$ is incident normally on a slit of width '$a$'. The distance between the slit and the screen is $2.5 \text{ m}$ and the distance of the second order maximum from the centre of the screen is $14.5 \text{ mm}$ in a diffraction pattern. The value of '$a$' is

$A$ single slit of width $0.20 \, mm$ is illuminated with light of wavelength $500 \, nm$. The observing screen is placed $80 \, cm$ from the slit. The width of the central bright fringe will be ..... $mm$.

In a single slit diffraction experiment,the width of the slit is made double its original width. Then the central maximum of the diffraction pattern will become

Light of wavelength $6000 \, \mathring{A}$ is incident on a slit of width $0.30 \, mm$. The screen is placed $2 \, m$ from the slit. Find the position of the first minima.

When the frequency of the light used is changed from $4 \times 10^{14} \ s^{-1}$ to $5 \times 10^{14} \ s^{-1}$, the angular width of the principal (central) maximum in a single slit Fraunhofer diffraction pattern changes by $0.6 \ \text{radian}$. What is the width of the slit? (Assume that the experiment is performed in vacuum.)