In Young's double slit experiment, the light of wavelength ' $\lambda$ ' is used. The intensity at a point on the screen is 'I' where the path difference is $\frac{\lambda}{4}$. If ' $\mathrm{I}_{\mathrm{o}}$ ' denotes the maximum intensity then the ratio of ' $\mathrm{I}_{\mathrm{o}}$ ' to ' I ' is $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)$

The intensity at a point on the screen in Young's double-slit experiment is given by $I = I_o \cos^2\left(\frac{\phi}{2}\right)$, where $I_o$ is the maximum intensity and $\phi$ is the phase difference.

The phase difference relates to path difference by $\phi = \frac{2\pi}{\lambda} \Delta x$. For $\Delta x = \frac{\lambda}{4}$, we have $\phi = \frac{\pi}{2}$.

Substituting into the intensity formula gives $I = I_o \cos^2\left(\frac{\pi}{4}\right)$. Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, this simplifies to $I = I_o \left(\frac{1}{2}\right)$.

The ratio is therefore $\frac{I_o}{I} = 2$, or equivalently $2:1$.

Final answer: $\boxed{\text{A}}$