1. Condition for minima in single-slit diffraction

For a single slit of width $a$, the minima occur at angles $\theta_m$ given by

$a\sin\theta_m = m\lambda, \quad m=1,2,3,\dots$

Here:

• second minimum to the left of central maximum corresponds to $m=2$ on the left,
• third minimum to the right corresponds to $m=3$ on the right.

2. Use the given angular separation

If the minima are on opposite sides of the central maximum, their angular separation is

$|\theta_{-2}| + |\theta_{+3}| = 30^\circ$

Let $\theta_2 + \theta_3 = 30^\circ$ with

$a\sin\theta_2 = 2\lambda, \qquad a\sin\theta_3 = 3\lambda$

3. Small-angle approximation

In diffraction problems of this type, minima are usually close to the central axis, so

$\sin\theta \approx \theta \quad (\text{in radians})$

Thus,

$\theta_2 \approx \frac{2\lambda}{a}, \qquad \theta_3 \approx \frac{3\lambda}{a}$

Therefore,

$\theta_2 + \theta_3 \approx \frac{5\lambda}{a}$

Given separation is $30^\circ = \frac{\pi}{6}$ rad, so

$\frac{5\lambda}{a} = \frac{\pi}{6}$

Hence,

$a = \frac{5\lambda}{\pi/6} = \frac{30\lambda}{\pi}$

4. Substitute $\lambda = 628\text{ nm}$

$a = \frac{30\times 628\times 10^{-9}}{\pi}\text{ m}$

$a \approx \frac{18840\times 10^{-9}}{3.1416}$

$a \approx 5.997\times 10^{-6}\text{ m}$

$a \approx 6\times 10^{-6}\text{ m} = 6\,\mu\text{m}$

5. Final answer

$\boxed{6}$

6. Comparison with stored answer

Stored correct answer = $6$

My derived answer also = $6$, so they agree.