In a single slit diffraction experiment, light of wavelength 600 nm is used and the first minimum is observed at an angle of 30°. The width of the slit is:

Correct option is D

​$$\text{Given:} \\\text{Wavelength of light } \lambda = 600 \, \text{nm} \\\text{The first minimum is observed at an angle } \theta = 30^\circ \\\text{We are asked to find the width of the slit.}$$

$$\text{For single-slit diffraction, the condition for the first minimum is:} \\d \sin \theta = n \lambda$$

$$\text{Where:} \\d \text{ is the width of the slit,} \\\theta \text{ is the angle for the first minimum,} \\n \text{ is the order of the minimum (for the first minimum, } n = 1), \\\lambda \text{ is the wavelength of light.}$$

$$\text{For the first minimum:} \\d \sin 30^\circ = 1 \times 600 \, \text{nm} \\\text{We know that } \sin 30^\circ = \frac{1}{2}, \text{ so:} \\d \times \frac{1}{2} = 600 \, \text{nm} \\d = 600 \times 2 = 1200 \, \text{nm} = 1.2 \, \mu\text{m}$$​​​​​