Answer
$\sqrt 2 :1$
Work Step by Step
Condition for the single slit diffraction minima is:
$a \sin\theta = m\lambda$
where;
$a$ be the slit width
$\lambda$ be the wavelength
It is given that;
the angle $\theta = 45^{\circ}$
and
for first order, m=1
Therefore, from the above formula we get ratio of slit to wavelength is:
$\frac{a}{\lambda}=\frac{m}{\sin \theta}$
$\frac{a}{\lambda}=\frac{1}{\sin 45^{\circ}}$
$\frac{a}{\lambda}=\frac{1}{(\frac{1}{\sqrt 2})}$
$a:\lambda=\sqrt 2 :1$
