To observe diffraction,the size of the obstacle

The condition for obtaining secondary maxima in the diffraction pattern due to a single slit is:

Light of wavelength $\lambda = 5000 \ \mathring{A}$ falls normally on a narrow slit. $A$ screen is placed at a distance of $1 \ m$ from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at $5 \ mm$ from the centre of the central maximum. The width of the slit is ....... $mm$.

In a single-slit diffraction experiment,the slit is illuminated by light of two wavelengths $\lambda_1$ and $\lambda_2$. It is observed that the $2^{nd}$ order diffraction minimum for $\lambda_1$ coincides with the $3^{rd}$ diffraction minimum for $\lambda_2$. Then:

$A$ 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:

$A$ parallel beam of light of wavelength $\lambda$ is incident normally on a single slit of width $d$. Diffraction bands are obtained on a screen placed at a distance $D$ from the slit. The second dark band from the central bright band will be at a distance given by