1. Identify the system

This is an Atwood machine with masses $m_1$ and $m_2$ connected by a light string over a smooth light pulley, with $m_2 > m_1$.

The acceleration of such a system is

$a = \frac{(m_2-m_1)g}{m_1+m_2}.$

Given:

$a = \frac{g}{\sqrt{2}}.$

So,

$\frac{(m_2-m_1)g}{m_1+m_2} = \frac{g}{\sqrt{2}}.$

Cancel $g$:

$\frac{m_2-m_1}{m_1+m_2} = \frac{1}{\sqrt{2}}.$

2. Express in terms of the ratio

Let

$r = \frac{m_1}{m_2}.$

Then $m_1 = rm_2$. Substitute:

$\frac{m_2-rm_2}{rm_2+m_2} = \frac{1}{\sqrt{2}}.$

$\frac{1-r}{1+r} = \frac{1}{\sqrt{2}}.$

3. Solve for $r$

Cross-multiplying:

$\sqrt{2}(1-r) = 1+r.$

Expand:

$\sqrt{2} - \sqrt{2}r = 1 + r.$

Bring $r$ terms together:

$\sqrt{2}-1 = r(\sqrt{2}+1).$

Hence,

$r = \frac{\sqrt{2}-1}{\sqrt{2}+1}.$

Therefore,

$\frac{m_1}{m_2} = \frac{\sqrt{2}-1}{\sqrt{2}+1}.$

4. Match with the options

This corresponds to Option C.

5. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.