In a single slit diffraction pattern, a light of wavelength $$6000$$ $$\mathring{A}$$ is used. The distance between the first and third minima in the diffraction pattern is found to be $$3$$ mm when the screen is placed $$50$$ cm away from slits. The width of the slit is ______ $$\times 10^{-4}$$ m.

In a single slit diffraction pattern, the position of the $$n$$th minimum is given by $$y_n = \frac{n\lambda D}{a}$$, where the wavelength $$\lambda = 6000$$ Angstrom $$= 6000 \times 10^{-10}$$ m $$= 6 \times 10^{-7}$$ m, the distance from slit to screen $$D = 50$$ cm $$= 0.50$$ m, and $$a$$ is the slit width.

The distance between the first and third minima is $$y_3 - y_1 = 3$$ mm $$= 3 \times 10^{-3}$$ m; substituting into the expression gives $$y_3 - y_1 = \frac{3\lambda D}{a} - \frac{1 \cdot \lambda D}{a} = \frac{2\lambda D}{a}.$$

Rearranging yields $$a = \frac{2\lambda D}{y_3 - y_1},$$ and thus $$a = \frac{2 \times 6 \times 10^{-7} \times 0.50}{3 \times 10^{-3}} = \frac{6 \times 10^{-7}}{3 \times 10^{-3}} = 2 \times 10^{-4} \text{ m}.$$

Expressed as $$\_\_\_ \times 10^{-4}$$ m, the numerical result is 2.