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 _______.

The angular width of the central maximum in a single slit diffraction pattern is given by $\Delta \theta = \dfrac{2\lambda}{a}$

Substituting the given values:
$\lambda = 628 \times 10^{-9}$ m
$a = 0.2 \times 10^{-3}$ m

$\Delta \theta = \dfrac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}}$ rad

$\Delta \theta = 6280 \times 10^{-6}$ rad $= 6.28 \times 10^{-3}$ rad

To convert radians to degrees, multiply by $\dfrac{180}{\pi}$:

$\Delta \theta = 6.28 \times 10^{-3} \times \dfrac{180}{3.14}$ degrees

$\Delta \theta = 2 \times 10^{-3} \times 180$ degrees

$\Delta \theta = 360 \times 10^{-3}$ degrees $= 36 \times 10^{-2}$ degrees

Comparing this with $\alpha \times 10^{-2}$ degrees, we get $\alpha = 36$.

Answer: $36$