Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are $$\theta_2$$ and $$\theta_1$$ respectively, $$(\theta_2 > \theta_1)$$. The temperature at the interface is:

A

$$\frac{\theta_2 + \theta_1}{2}$$

B

$$\frac{\theta_1}{6} + \frac{5\theta_2}{6}$$

C

$$\frac{\theta_1}{3} + \frac{2\theta_2}{3}$$

D

$$\frac{\theta_1}{10} + \frac{9\theta_2}{10}$$