If $I_0$ is the intensity of the principal maximum in the single slit diffraction pattern,then what will be the intensity when the slit width is doubled?

The distance between the first and the sixth minima in the diffraction pattern of a single slit is $0.5 \, mm$. The screen is $0.5 \, m$ away from the slit. If the wavelength of light used is $5000 \, \mathring{A}$,then the slit width will be ....... $mm$.

The Fraunhofer diffraction pattern of a single slit is formed in the focal plane of a lens of focal length $1\,m$. The width of the slit is $0.3\,mm$. If the third minimum is formed at a distance of $5\,mm$ from the central maximum, then the wavelength of light will be......$\mathring{A}$

$A$ diffraction pattern is obtained by making blue light incident on a narrow slit. If blue light is replaced by red light,then:

Two wavelengths $590 \ nm$ and $596 \ nm$ of sodium light are used one after another to study the diffraction taking place at a single slit of aperture $2.4 \ mm$. The distance between the slit and the screen is $2 \ m$. The separation between the positions of the first secondary maximum of the diffraction pattern obtained in the two cases is:

If $\lambda = 6000 \, \mathring{A}$ and $a = 18 \times 10^{-5} \, \text{cm}$,find the angular width of the central maximum in degrees $(^\circ)$.