1. Condition for minima in single-slit diffraction

For a single slit of width $a$, minima occur at

$a\sin\theta = m\lambda$

where $m=1,2,3,\dots$

For the second order minimum, we take

$m=2$

so,

$a\sin\theta = 2\lambda$

2. Given data

• Wavelength: $\lambda = 600\,\text{nm} = 600\times 10^{-9}\,\text{m}$
• Slit width: $a = 0.4\,\text{mm} = 0.4\times 10^{-3}\,\text{m}$

3. Find angular position of second minimum

Using small-angle approximation, $\sin\theta \approx \theta$:

$\theta = \frac{2\lambda}{a}$

Substitute values:

$\theta = \frac{2\times 600\times 10^{-9}}{0.4\times 10^{-3}}$

$\theta = \frac{1200\times 10^{-9}}{0.4\times 10^{-3}}$

$\theta = 3\times 10^{-3}\,\text{rad}$

4. Angular divergence corresponding to second order minima

The angular divergence means the angle between the minima on both sides of the central axis:

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

$\Delta\theta = 6\times 10^{-3}\,\text{rad}$

So the required number is

$\boxed{6}$