Plane microwaves are incident on a long slit of width 5.0 cm ....

The condition for the diffraction minima in a single slit experiment is given by $a \sin \theta = n \lambda$, where $a$ is the slit width, $\theta$ is the angle of diffraction, $n$ is the order of the minimum, and $\lambda$ is the wavelength.

For the first diffraction minimum, $n = 1$.

Given $a = 5.0 \text{ cm}$ and \theta = 30^{}^{{}^{\circ}}.

Substituting the values into the formula:

5.0 \sin 30^{}^{{}^{\circ}} = 1 \times \lambda

$\lambda = 5.0 \times \dfrac{1}{2}$

$\lambda = 2.5 \text{ cm}$