1. Condition for minima in single-slit diffraction

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

$b\sin\theta_n = n\lambda, \qquad n=1,2,3,\dots$

For small angles,

$\sin\theta_n \approx \tan\theta_n \approx \frac{y_n}{D}$

so the position of the $n$-th minimum on the screen is

$y_n = \frac{n\lambda D}{b}$

Thus, minima are equally spaced.

2. Use the given data

Given:

• second minimum at $y_2 = 3\text{ cm}$
• fourth minimum at $y_4 = 6\text{ cm}$
• screen distance $D=1\text{ m}$

From the formula,

$y_n = n\left(\frac{\lambda D}{b}\right)$

So,

$y_2 = 2\left(\frac{\lambda D}{b}\right)=3\text{ cm}$

Hence,

$\frac{\lambda D}{b} = 1.5\text{ cm}$

This is the distance of the first minimum from the central maximum:

$y_1 = 1.5\text{ cm}$

This also agrees with the fourth minimum:

$y_4 = 4y_1 = 4(1.5)=6\text{ cm}$

3. Width of the central maximum

The central maximum extends from the first minimum on one side to the first minimum on the other side.

Therefore, width of central maximum

$= 2y_1 = 2(1.5\text{ cm}) = 3.0\text{ cm}$

4. Evaluate options

• A: $1.5\text{ cm}$ $\to$ this is only the distance from center to first minimum, not full central width.
• B: $3.0\text{ cm}$ $\to$ correct.
• C: $4.5\text{ cm}$ $\to$ incorrect.
• D: $6.0\text{ cm}$ $\to$ incorrect.

Final Answer: $\boxed{3.0\text{ cm}}$ (Option B)