A slit of width $10 \times 10^{-7} \mathrm{~m}$ is illuminated by light of wavelength $500 \mathrm{~nm}$. Angular position of the first minimum is

The condition for the first minimum in a single-slit diffraction pattern is given by $a \sin \theta = n \lambda$, where $a$ is the slit width, $\theta$ is the angular position, $n$ is the order of the minimum, and $\lambda$ is the wavelength of light.

Given values are $a = 10 \times 10^{-7} \text{ m} = 10^{-6} \text{ m}$, $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5 \times 10^{-7} \text{ m}$, and $n = 1$ for the first minimum.

Substituting these values into the formula:

$10^{-6} \sin \theta = 1 \times 5 \times 10^{-7}$

$\sin \theta = \dfrac{5 \times 10^{-7}}{10^{-6}} = 0.5$

Since $\sin \theta = 0.5$, the angle $\theta$ is $30^{\circ}$.

Answer: 30$^\circ$