Question

Assume (unrealistically) that a TV station acts as a point source broadcasting isotopically at $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{MW}}$. What is the intensity of the transmitted signal reaching Proxima Centauri, the star nearest our solar system,$\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{\text{ly}}$away? (An alien civilization at that distance might be able to watchXFiles.) A light-year ($\mathit{l}\mathit{y}$) is the distance light travels in one year.

Short answer

The intensity of the transmitted signal reaching Proxima Centauri, the star nearest our solar system $4.3 \text{ly}$ is,$4.8\times {10}^{-29}\mathrm{W}/{\mathrm{m}}^2$ .

Step-by-step solution

Listing the given quantities

Source power

$\begin{array}{rcl}{P}_s& =& 1.0 \text{MW}\\ & =& 1.0\times {10}^6 \text{W}\end{array}$

Distance of the star is, $r=4.3 \text{ly}$

Step 2:Understanding the concepts of intensity

The source is emitting power isotopically. The received power intensity depends on the source power and the distance.

Formula:

$I=\frac{P_s}{4\pi R^2}$

Calculations of the intensity of the transmitted signal reaching Proxima Centauri, the star nearest our solar system,  4.3ly away

The distance is given in light years, by converting it in to meters:

$1 \text{ly}=9.46\times {10}^{15} \text{m}$

$\begin{array}{rcl}4.3 \text{ly}& =& 4.3\times 9.46\times {10}^{15} \text{m}\\ & =& 4.068\times {10}^{16} \text{m}\end{array}$

Using the intensity formula:

$I=\frac{P_s}{4\pi R^2}$

Substitute all the value in the above equation.

$\begin{array}{rcl}I& =& \frac{1.0\times {10}^{6}W}{4\mathrm{\pi }\times {\left(4.068\times {10}^{16}\mathrm{m}\right)}^2}\\ & =& 4.8\times {10}^{-29}\mathrm{W}/{\mathrm{m}}^2\end{array}$

The intensity of the transmitted signal reaching Proxima Centauri, the star nearest our solar system, 4.3lyaway $4.8\times {10}^{-29}\mathrm{W}/{\mathrm{m}}^2$