What happens if we close one slit in a Young's double-slit experiment?

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

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:

The condition for diffraction is

In a single slit diffraction of light of wavelength $\lambda$ by a slit of width $e$,the size of the central maximum on a screen at a distance $b$ is

In a single slit diffraction experiment, a parallel beam of green light of wavelength $\lambda = 550 \,nm$ passes through a slit of width $a = 0.20 \,mm$. The transmitted light is collected on a screen at a distance $D = 100 \,cm$ away. The distance of the first-order minima from the central maximum is $x \times 10^{-5} \,m$. The value of $x$ is: