The electric field between the plates of a capacitor in vacuum (or air) is $E_0$. When a dielectric slab of dielectric constant $K$ is introduced, the electric field inside the dielectric decreases to $E = \frac{E_0}{K}$ [NCERT Class 12, Sec 2.10].

1. Inverse Relationship: The electric field is inversely proportional to the dielectric constant ($E \propto 1/K$).
2. Comparison: We are given $K_1 < K_2$. Therefore, $\frac{1}{K_1} > \frac{1}{K_2}$.
3. Field Magnitude: This implies that the electric field inside the first slab ($E_1 = E_0/K_1$) is greater than the electric field inside the second slab ($E_2 = E_0/K_2$).
4. Graph Analysis: The correct graph should show regions of constant electric field. The field will be highest in the air regions ($E_0$), intermediate in the slab with $K_1$, and lowest in the slab with $K_2$. Thus, the graph steps down from $E_0$ to $E_1$, goes back to $E_0$ (if there is an air gap), and then steps down to a lower value $E_2$ for the second slab.