The outer surface of star in the form of a sphere radiates heat as a black body at temperature ' $T$ '. The…

The outer surface of star in the form of a sphere radiates heat as a black body at temperature ' $T$ '. The total radiant energy per unit area, normal to the direction of incidence, received at a distance ' R ' from the centre of a star of radius ' r ' is ( $\mathrm{R}>\mathrm{r}$ ) ( $\sigma=$ Stefan's constant)
  1. $\frac{\sigma r^2 \mathrm{~T}^4}{\mathrm{R}^2}$
  2. $\frac{\sigma r^2 T^4}{4 \pi R^2}$
  3. $\frac{\sigma \mathrm{r}^2 \mathrm{~T}^4}{\mathrm{R}^4}$
  4. $\frac{4 \pi \sigma r^2 T^4}{R^2}$

Solution

The radiant energy per unit area received normal to incidence at distance $R$ from a star of radius $r$ radiating as a black body at temperature $T$ is determined through the Stefan-Boltzmann law and spherical radiation geometry.

The star's total radiative power is $P = \sigma (4\pi r^2) T^4$.

At distance $R$, this power distributes uniformly over the spherical surface area $4\pi R^2$, yielding the intensity $I = \frac{P}{4\pi R^2}$.

Substituting gives $I = \frac{\sigma (4\pi r^2) T^4}{4\pi R^2} = \frac{\sigma r^2 T^4}{R^2}$.

Final Answer: $\boxed{\text{A}}$

Asked in: MHT CET 2025 (05 May Shift 2)

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