Condition for diffraction maximum is $a \sin \theta=(2 n+1) \frac{\pi}{2}$

For first Maxima, $n=1, a \sin \theta=\frac{3 \lambda}{2}$

Since $\theta$ is very less, so

$$ \begin{aligned} & \sin \theta \approx \tan \theta \\\\ & \therefore a \tan \theta=\frac{3 \lambda}{2} \end{aligned} $$

$$ \begin{aligned} a\left(\frac{y}{D}\right) =\frac{3 \lambda}{2} \\\\ \Rightarrow y =\frac{3 \lambda \mathrm{D}}{2 a} \end{aligned} $$

For the two cases given

$$ \begin{aligned} y_2-y_1 & =\frac{3 \mathrm{D}}{2 a}\left(\lambda_2-\lambda_1\right) \\\\ & =\frac{3}{2} \times \frac{2}{0 \cdot 5 \times 10^{-3}}(655-650) \times 10^{-9} \\\\ & =6 \times 10^3 \times 5 \times 10^{-9} \\\\ & =3 \times 10^{-5} \mathrm{~m} \end{aligned} $$