$A$ plane wavefront of wavelength $6000 \, \mathring A$ is incident on a slit of width $0.2 \, mm$. $A$ Fraunhofer diffraction pattern is obtained on a screen at a distance of $2 \, m$ from the slit. What is the width of the central maximum in $mm$?

In a Fraunhofer diffraction of a single slit, when a slit is illuminated by a light of wavelength $6480 \text{ Å}$, the angular width of the central maximum is measured. When the slit is illuminated by light of another wavelength '$\lambda$',the angular width decreases by $25\%$. The value of '$\lambda$' in $\text{Å}$ is:

$A$ light beam of wavelength $800 \,nm$ passes through a single slit and is projected on a screen kept at $5 \,m$ away from the slit. What should be the slit width for the ray optics approximation to be valid (in $\,mm$)?

Consider the following statements $A$ and $B$ and identify the correct answer:
$A$. Fresnel's diffraction pattern occurs when the source of light or the screen on which the diffraction pattern is seen, or when both are at a finite distance from the aperture.
$B$. Diffracted light can be used to estimate the helical structure of nucleic acids.

In Fraunhofer diffraction pattern, slit width is $0.2 \, mm$ and screen is at $2 \, m$ away from the lens. If wavelength of light used is $5000 \, \mathring{A}$, then the distance between the first minimum on either side of the central maximum is (assume $\theta$ is small and measured in radians).