$A$ parallel beam of monochromatic light of wavelength $600 \,nm$ passes through a single slit of $0.4 \,mm$ width. The angular divergence corresponding to the second-order minima would be $...... \times 10^{-3} \,rad$.

In the case of Fraunhofer diffraction due to a single slit,the intensity of the central maximum is $I_0$. If the width of the slit is doubled,what will be the new intensity of the central maximum?

In a single slit diffraction experiment,when the distance of separation between the slit and the screen is doubled,the angular separation between the fringes

The phenomenon of diffraction of light was discovered by

In a single slit diffraction experiment, a parallel beam of green light of wavelength $\lambda = 550 \,nm$ passes through a slit of width $a = 0.20 \,mm$. The transmitted light is collected on a screen at a distance $D = 100 \,cm$ away. The distance of the first-order minima from the central maximum is $x \times 10^{-5} \,m$. The value of $x$ 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: