Two long concentric black cylinders of radii R_i (inner) and Rₒ (outer) are in vacuum and radiate energy according to their temperatures. What fraction of the radiation emitted inward by the outer cylinder eventually falls back on the outer cylinder itself?

1 - R_i/Rₒ
1 - sqrt(R_i/Rₒ)
sqrt(1 - R_i/Rₒ)
1 - (R_i/Rₒ)^(1/3)

Correct answer: 1 - R_i/Rₒ

Solution

By the view-factor relation, a fraction R_i/Rₒ of the outer cylinder's inward radiation strikes the inner cylinder, so the remaining 1 - R_i/Rₒ returns to the outer cylinder.

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