The view factor $F_{ij}$ represents the fraction of the radiation leaving surface $i$ that strikes surface $j$. For concentric spheres, there is a special relationship due to symmetry. The view factor $F_{11}$ represents the fraction of radiation leaving the inner sphere that strikes itself. For any surface, the sum of view factors to all other surfaces must be equal to 1. Thus, $F_{11} + F_{12} = 1$, where $F_{12}$ is the fraction of radiation leaving the inner sphere (surface 1) that strikes the outer sphere (surface 2). Since the inner sphere is completely enclosed by the outer sphere, all radiation leaving the inner sphere must strike the outer sphere, so $F_{12} = 1$. Therefore, $F_{11} = 1 - F_{12} = 1 - 1 = 0$.Now, consider the view factor $F_{22}$, which represents the fraction of radiation leaving the outer sphere (surface 2) that strikes itself. The general relationship between view factors is $A_i F_{ij} = A_j F_{ji}$. Here, $A_1$ is the surface area of the inner sphere, $A_1 = 4\pi r_1^2$. And $A_2$ is the surface area of the outer sphere, $A_2 = 4\pi r_2^2$. Using the reciprocity relation for $F_{12}$: $A_1 F_{12} = A_2 F_{21}$. We know $F_{12} = 1$ (since the outer sphere completely surrounds the inner sphere). So, $A_1 \times 1 = A_2 F_{21}$. $4\pi r_1^2 \times 1 = 4\pi r_2^2 F_{21}$. $r_1^2 = r_2^2 F_{21}$. $F_{21} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.Now, we need to find $F_{22}$. The sum of view factors from surface 2 to all surfaces must be 1. Surface 2 sees itself ($F_{22}$) and the inner sphere (surface 1, $F_{21}$). So, $F_{22} + F_{21} = 1$. $F_{22} = 1 - F_{21}$. Substituting the value of $F_{21}$: $F_{22} = 1 - \left(\frac{r_1}{r_2}\right)^2$.