1. Given data

• Wavelengths: $\lambda_1 = 650\ \text{nm} = 650 \times 10^{-9}\ \text{m}$ $\lambda_2 = 655\ \text{nm} = 655 \times 10^{-9}\ \text{m}$
• Slit width: $a = 0.5\ \text{mm} = 0.5 \times 10^{-3}\ \text{m}$
• Distance to screen: $D = 2.0\ \text{m}$

We need the separation between the first maxima for the two wavelengths.

2. Position of first secondary maximum in single-slit diffraction

For single-slit diffraction, $I = I_0\left(\frac{\sin \beta}{\beta}\right)^2, \qquad \beta = \frac{\pi a \sin\theta}{\lambda}$

Secondary maxima occur approximately at $\beta = \frac{3\pi}{2},\ \frac{5\pi}{2}, \dots$

For the first secondary maximum, $\beta = \frac{3\pi}{2}$

So, $\frac{\pi a \sin\theta}{\lambda} = \frac{3\pi}{2}$ $a\sin\theta = \frac{3\lambda}{2}$

For small angles, $y \approx D\tan\theta \approx D\sin\theta$

Hence position of first maximum: $y = D\frac{3\lambda}{2a}$

3. Separation of first maxima for the two wavelengths

$\Delta y = D\frac{3}{2a}(\lambda_2 - \lambda_1)$

Now, $\lambda_2 - \lambda_1 = (655-650)\times 10^{-9} = 5\times 10^{-9}\ \text{m}$

Thus, $\Delta y = 2 \cdot \frac{3}{2\times 0.5\times 10^{-3}} \cdot 5\times 10^{-9}$

First simplify: $2 \cdot \frac{3}{2\times 0.5\times 10^{-3}} = 2 \cdot \frac{3}{10^{-3}} \cdot \frac{1}{1}?$ Since $2\times 0.5\times 10^{-3} = 10^{-3}$ we get $\Delta y = 2 \cdot \frac{3}{10^{-3}} \cdot 5\times 10^{-9}$ $= 2 \cdot 3 \cdot 5 \times 10^{-6}$ $= 30 \times 10^{-6}$ $= 3 \times 10^{-5}\ \text{m}$

4. Required integer

The separation is $3 \times 10^{-5}\ \text{m}$

So the blank is: $\boxed{3}$