The first diffraction minimum due to the single slit diffraction is seen at $$\theta = 30°$$ for a light of wavelength $$5000\text{\AA}$$ falling perpendicularly on the slit. The width of the slit is

For a single slit diffraction, the position of minima is given by:

$$a \sin\theta = n\lambda$$

Given:

First minimum, $$n = 1$$

Angle of diffraction, $$\theta = 30^\circ$$

Wavelength of light, $$\lambda = 5000\text{ \AA} = 5000 \times 10^{-10}\text{ m} = 5 \times 10^{-7}\text{ m}$$

Substituting the values into the formula:

$$a \sin 30^\circ = 1 \times 5 \times 10^{-7}$$

$$a \times \frac{1}{2} = 5 \times 10^{-7}$$

$$a = 2 \times 5 \times 10^{-7}\text{ m}$$

$$a = 10^{-6}\text{ m} = 1\text{ }\mu\text{m}$$