$A$ 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 $mm$?

The angular width of the central maxima in Fraunhofer diffraction for $\lambda = 6000 \; \mathring{A}$ is $\theta_{0}$. When the same slit is illuminated by another monochromatic light,the angular width decreases by $30 \%$. The wavelength of this light is ....... $\mathring{A}$.

In the Fraunhofer diffraction experiment,$L$ is the distance between the screen and the obstacle,$b$ is the size of the obstacle,and $\lambda$ is the wavelength of the incident light. The general condition for the applicability of Fraunhofer diffraction is:

The angular separation of the central maximum in the Fraunhofer diffraction pattern is measured. The slit is illuminated by light of wavelength $6000 Å$. If the slit is illuminated by light of another wavelength,the angular separation decreases by $20 \%$. The wavelength of light used is (in $Å$)

The direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by ($a$ is the width of the slit):

In a Fraunhofer diffraction,light of wavelength $\lambda$ is incident on a slit of width $d$. The diffraction pattern is observed on a screen placed at a distance $D$. If the linear width of the central maximum is equal to two times the width of the slit,then the value of $D$ is: