A single slit of width $$a$$ is illuminated by a monochromatic light of wavelength $$600$$ nm. The value of $$a$$ for which first minimum appears at $$\theta = 30°$$ on the screen will be :

For single slit diffraction, the condition for first minimum is:

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

Given: $$\lambda = 600$$ nm, $$\theta = 30°$$

$$a = \frac{\lambda}{\sin\theta} = \frac{600 \text{ nm}}{\sin 30°} = \frac{600}{0.5} = 1200 \text{ nm} = 1.2 \ \mu\text{m}$$

This matches option 1: 1.2 μm.

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