Chapter 33, Problem 16P

To determine

The relative distance between both the stars from Earth.

Answer to Problem 16P

The star with the brightness at $400\text{ nm}$ is 4 times farther away than the star with the brightness at $800\text{ nm}$ .

Explanation of Solution

Given:

Intensity of two stars peaks at $400\text{ nm}$ and $800\text{ nm}$ , and of same size.

Formula used:

According to Wien’s law,

  ${\lambda }_{\text{P}}\text{t =}\alpha$

According to Stefan Boltzmann equation,

  $\text{P = }\beta {\text{At}}^4$

Calculation:

Using Wien’s law, the relation between temperatures is obtained as

  $\begin{array}{l}{\lambda }_{\text{P1}}{\text{T}}_1\text{ =}{\lambda }_{\text{P2}}{\text{T}}_2\\ \frac{{\text{T}}_2}{{\text{T}}_1}=\frac{{\lambda }_{\text{P1}}}{{\lambda }_{\text{P2}}}=\frac{800\text{ nm}}{400\text{ nm}}=2\end{array}$

Now, using the Stefan Boltzmann equation,

  $\begin{array}{l}{\text{l}}_1={\text{l}}_2\Rightarrow \frac{{\text{L}}_1}{4\pi {\text{d}}_1^2}=\frac{{\text{L}}_2}{4\pi {\text{d}}_2^2}\\ \frac{{\text{d}}_2^2}{{\text{d}}_1^2}=\frac{{\text{L}}_2}{{\text{L}}_1}=\frac{{\text{P}}_2}{{\text{P}}_1}=\frac{\beta {\text{A}}_2{\text{T}}_2^4}{\beta {\text{A}}_1{\text{T}}_1^4}=\frac{4\pi {\text{r}}_2^2{\text{T}}_2^4}{4\pi {\text{r}}_1^2{\text{T}}_1^4}=\frac{{\text{T}}_2^4}{{\text{T}}_1^4}\\ \frac{{\text{d}}_2}{{\text{d}}_1}={\left(\frac{{\text{T}}_2}{{\text{T}}_1}\right)}^2={\left(2\right)}^2=4\end{array}$

Conclusion:

The star with the brightness at $400\text{ nm}$ is 4 times farther away than the star with the brightness at $800\text{ nm}$ .