Answer
$$\frac{d_{2}}{d_{1}}=2.8$$
$\therefore$ The star with the peak at $450$ $\mathrm{nm}$ is equal to $2.8$ times farther away than the star with the peak at $750$ $\mathrm{nm}$ .
Work Step by Step
$\text { the following information are given }$
$$ b_{1}=b_{2}, r_{1}=r_{2}, \lambda_{\mathrm{Pl}}=750 \mathrm{nm}, \text { and } \lambda_{\mathrm{P} 2}=450 \mathrm{nm} . $$
$\text {from Wien's law (Eq. } 27-1) \text{we know that} $
$ \lambda_{\mathrm{P}} T=\alpha, \text { where } \alpha \text { is a constant, so } \lambda_{\mathrm{Pl}} T_{1}=\lambda_{\mathrm{P} 2} T_{2} $
$\text {The Stefan-Boltzmann equation (Eq. } 14-6) \text{ says that the power output}
\\ \text{ of a star is given by}$
$$P=\beta A T^{4},$$ where $\beta$ is a constant and $A$ is the radiating area.
$\therefore$
$$\lambda_{\mathrm{P} 1} T_{1}=\lambda_{\mathrm{P} 2} T_{2} \quad \rightarrow \quad \frac{T_{2}}{T_{1}}=\frac{\lambda_{\mathrm{P} 1}}{\lambda_{\mathrm{P} 2}}$$
$$b_{1}=b_{2} \rightarrow \frac{L_{1}}{4 \pi d_{1}^{2}}=\frac{L_{2}}{4 \pi d_{2}^{2}} \rightarrow $$$$\frac{d_{2}^{2}}{d_{1}^{2}}=\frac{L_{2}}{L_{1}}=\frac{P_{2}}{P_{1}}=\frac{\beta A_{2} T_{2}^{4}}{\beta A_{1} T_{1}^{4}}=$$$$=\frac{4 \pi r_{2}^{2} T_{2}^{4}}{4 \pi r_{1}^{2} T_{1}^{4}}=\frac{T_{2}^{4}}{T_{1}^{4}}=\left(\frac{T_{2}}{T_{1}}\right)^{4} \rightarrow$$
$$\frac{d_{2}}{d_{1}}=\left(\frac{T_{2}}{T_{1}}\right)^{2}=\left(\frac{\lambda_{\mathrm{P} 1}}{\lambda_{\mathrm{P} 2}}\right)^{2}=\left(\frac{750}{450}\right)^{2}=2.8$$
$\therefore$ The star with the peak at $450$ $\mathrm{nm}$ is equal to $2.8$ times farther away than the star with the peak at $750$ $\mathrm{nm}$ .
