The magnitude of the electric field vector at the wall is
$\begin{align*}E_0 \frac{\sin \left(\beta/2\right)}{\beta/2}\end{align*}$
where
$\begin{align*}\beta = \frac{2 \pi}{\lambda} d \sin \theta\end{align*}$
The minima occur when
$\begin{align*}\sin \left(\beta/2\right) = 0\end{align*}$
which implies that the first minimum occurs when
$\begin{align*}\beta/2 = \pi\end{align*}$
This implies that
$\begin{align*}2 \pi = \frac{2 \pi}{\lambda} d \sin \theta \Rightarrow d = \lambda/\sin \theta\end{align*}$
If we use the small angle approximation for $\theta$ , then we find that
$\begin{align*}d = \frac{\lambda}{\theta} = \frac{400 \cdot 10^{-9} \mbox{ m}}{4 \cdot 10^{-3}} = 100 \cdot 10^{-6} \mbox{ m} ...$
Therefore, answer (C) is correct.

Solution to 1996 Problem 57