$A$ parallel monochromatic beam of light 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 first maximum of the diffraction pattern,the phase difference between the rays coming from the edges of the slit is

$A$ screen is placed at $50 \ cm$ from a single slit,which is illuminated with light of wavelength $600 \ nm$. If the separation between the $1^{st}$ and $3^{rd}$ minima in the diffraction pattern is $3 \ mm$,then the slit width is: (in $mm$)

In a Fraunhofer diffraction experiment,a single slit of width $0.5 ~mm$ is illuminated by a monochromatic light of wavelength $600 ~nm$. The diffraction pattern is observed on a screen at a distance of $50 ~cm$ from the slit. What will be the linear separation of the first order minima (in $~mm$)?

The diffraction pattern of light of wavelength $400 \ nm$ diffracting from a slit of width $0.2 \ mm$ is focused on the focal plane of a convex lens of focal length $100 \ cm$. The width of the $1^{\text{st}}$ secondary maxima will be:

$A$ linear aperture whose width is $0.02\, cm$ is placed immediately in front of a lens of focal length $60\, cm.$ The aperture is illuminated normally by a parallel beam of wavelength $5 \times 10^{-5}\, cm.$ The distance of the first dark band of the diffraction pattern from the centre of the screen is .....$cm$

The box of a pinhole camera,of length $L$,has a hole of radius $a$. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $\lambda$,the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $b_{min}$) when