In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0.2 mm, the angular width of central maximum is $$\alpha \times 10^{-2}$$ degrees. The value of $$\alpha$$ is __________.

For a single-slit of width $$a$$, the first diffraction minimum occurs at $$a \sin\theta = \lambda$$. For small angles in diffraction experiments $$\sin\theta \approx \theta$$ (in radians).

Given: wavelength $$\lambda = 628\ \text{nm} = 628 \times 10^{-9}\ \text{m}$$ slit width $$a = 0.2\ \text{mm} = 0.2 \times 10^{-3}\ \text{m} = 2 \times 10^{-4}\ \text{m}$$

First minimum angle: $$\theta = \frac{\lambda}{a} = \frac{628 \times 10^{-9}}{2 \times 10^{-4}}$$ $$\theta = 314 \times 10^{-7} = 3.14 \times 10^{-3}\ \text{rad}$$

The central maximum extends from $$-\theta$$ to $$+\theta$$, so its angular width is $$2\theta = 2 \times 3.14 \times 10^{-3} = 6.28 \times 10^{-3}\ \text{rad}$$

Convert radians to degrees using $$1\ \text{rad} = \frac{180}{\pi} \approx 57.3^{\circ}$$: $$\text{width in degrees} = 6.28 \times 10^{-3} \times 57.3 \approx 0.36^{\circ}$$

Write this as $$0.36^{\circ} = 36 \times 10^{-2}\ ^{\circ}$$, hence $$\alpha = 36$$.

Final Answer: 36