A parallel beam of monochromatic light of wavelength $$600 \text{ nm}$$ passes through single slit of $$0.4 \text{ mm}$$ width. Angular divergence corresponding to second order minima would be ______ $$\times 10^{-3} \text{ rad}$$.

Find the angular divergence for the second-order minima in single-slit diffraction.

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

where $$a$$ is the slit width, $$n$$ is the order, and $$\lambda$$ is the wavelength.

$$\sin\theta = \frac{2\lambda}{a} = \frac{2 \times 600 \times 10^{-9}}{0.4 \times 10^{-3}} = \frac{1200 \times 10^{-9}}{4 \times 10^{-4}} = 3 \times 10^{-3}$$

Since $$\sin\theta$$ is very small, $$\theta \approx \sin\theta = 3 \times 10^{-3}$$ rad.

The angular divergence is the total angle between the 2nd order minima on both sides of the central maximum:

$$\text{Angular divergence} = 2\theta = 2 \times 3 \times 10^{-3} = 6 \times 10^{-3} \text{ rad}$$

The correct answer is 6.