As per the given figure, two plates $A$ and $B$ of thermal conductivity $K$ and $2K$ are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is $120 \mathrm {~cm} ^ { 2 }$ for each plate. The equivalent thermal conductivity of the compound plate is $\left( 1 + \frac { 5 } { \alpha } \right) K$, then the value of $\alpha$ will be $\_\_\_\_$ .