As the given figure of two slit interference arrangements, we can write
$$
\begin{gathered}
\mathrm{T}_2 \mathrm{P}=\mathrm{T}_2 \mathrm{O}+\mathrm{OP}=\mathrm{D}+\mathrm{x} \\
(\because \mathrm{OP}=\mathrm{x})
\end{gathered}
$$
and
$$
\begin{aligned}
\mathrm{T}_1 \mathrm{P} &=\mathrm{T}_1 \mathrm{O}-\mathrm{OP}=\mathrm{D}-\mathrm{x} \\
\mathrm{S}_1 \mathrm{P} &=\sqrt{\left(\mathrm{S}_1 \mathrm{~T}_1\right)^2+\left(\mathrm{PT}_1\right)^2} \\
&=\sqrt{\mathrm{D}^2+(\mathrm{D}-\mathrm{x})^2} \\
\mathrm{~S}_2 \mathrm{P} &=\sqrt{\left(\mathrm{S}_2 \mathrm{~T}_2\right)^2+\left(\mathrm{T}_2 \mathrm{P}\right)^2} \\
&=\sqrt{\mathrm{D}^2+(\mathrm{D}+\mathrm{x})^2}
\end{aligned}
$$
and
So, path difference $=\left(\mathrm{S}_2 \mathrm{P}-\mathrm{S}_1 \mathrm{P}\right)$
The minima will occur when
$$
\mathrm{S}_2 \mathrm{P}-\mathrm{S}_1 \mathrm{P}=(2 \mathrm{n}-1) \frac{\lambda}{2}
$$
For first minima $\mathrm{P}=\frac{\lambda}{2}$ at $\mathrm{n}=1$
So, $\left[D^2+(D+x)^2\right]^{1 / 2}-\left[D^2+(D-x)^2\right]^{1 / 2}=\frac{\lambda}{2}$
If $(x=D)$ then,
We can write $\left[\mathrm{D}^2+4 \mathrm{D}^2\right]^{1 / 2}-\left[\mathrm{D}^2+0\right]^{1 / 2}=\frac{\lambda}{2}$
$$
\begin{gathered}
{\left[5 \mathrm{D}^2\right]^{1 / 2}-\left[\mathrm{D}^2\right]^{1 / 2}=\frac{\lambda}{2}} \\
\sqrt{5} \mathrm{D}-\mathrm{D}=\frac{\lambda}{2} \\
\mathrm{D}(\sqrt{5}-1)=\frac{\lambda}{2} \\
\text { or } \quad \mathrm{D}=\frac{\lambda}{2(\sqrt{5}-1)} \times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{3.236 \lambda}{2 \times 4} \\
\mathrm{D}=\frac{3.236}{8} \lambda=0.404 \lambda
\end{gathered}
$$