1. Condition for first minimum in single-slit diffraction

For a single slit of width $a$, the minima are given by:

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

For the first minimum, $m=1$, so:

$a\sin\theta = \lambda$

2. Substitute the given values

Given:

• $\lambda = 600\,\text{nm} = 0.6\,\mu\text{m}$
• $\theta = 30^\circ$

Using $\sin 30^\circ = \frac{1}{2}$:

$a \cdot \frac{1}{2} = 0.6\,\mu\text{m}$

3. Solve for $a$

$a = \frac{0.6}{1/2} = 1.2\,\mu\text{m}$

4. Match with the options

$a = 1.2\,\mu\text{m}$

So the correct option is:

D: $1.2\,\mu\text{m}$

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

Hence, the derived answer agrees with the stored answer.