1. Given data

• Slit separation: $d = 0.6\text{ mm} = 0.6 \times 10^{-3}\text{ m}$
• Screen distance: $D = 80\text{ cm} = 0.8\text{ m}$
• The first dark fringe is observed directly opposite to one of the slits.

2. Interpret the condition

In YDSE, if the point directly opposite to one slit is considered, its distance from the central line is:

$y = \frac{d}{2}$

This is because the central bright fringe lies opposite the midpoint of the two slits, so each slit is at a transverse distance $d/2$ from the center.

Thus,

$y = \frac{0.6\times 10^{-3}}{2} = 0.3\times 10^{-3}\text{ m}$

3. Condition for first dark fringe

Position of the first dark fringe is:

$y_1 = \frac{\lambda D}{2d}$

Given that this first dark fringe occurs at $y = d/2$, so:

$\frac{\lambda D}{2d} = \frac{d}{2}$

Multiply both sides by $2$:

$\frac{\lambda D}{d} = d$

So,

$\lambda = \frac{d^2}{D}$

4. Substitute values

$\lambda = \frac{(0.6\times 10^{-3})^2}{0.8}$

$= \frac{0.36\times 10^{-6}}{0.8}$

$= 0.45\times 10^{-6}\text{ m}$

$= 4.5\times 10^{-7}\text{ m}$

5. Convert to nm

$1\text{ nm} = 10^{-9}\text{ m}$

So,

$\lambda = 4.5\times 10^{-7}\text{ m} = 450\text{ nm}$

6. Comparison with stored answer

Derived answer = $450$

Stored correct answer = $450$

They match.