In Young's double slit experiment, the light of wavelength ' $\lambda$ ' is used. The intensity at a point on the screen is ' T ' where the path difference is $\lambda \frac{-}{4}$. If ' $\mathrm{I}_0$ ' denotes the maximum intensity then the ratio of ' $\mathrm{I}_0$ ' to ' I ' is $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)$

Two cells of e.m.f.s $E_1$ and $E_2\left(E_1>E_2\right)$ are connected as shown in figure.

When the potentiometer is connected between A and B , the balancing length of the potentiometer wire is 3.60 m . On connecting the potentiometer between A and C , the balancing length is 0.90 m . The ratio $E_1 / E_2$ is