Calculate Power Radiated by a Star in Specific Spectral Range

We can use the Stefan-Boltzmann law to calculate the total power radiated by the star:

$P = \sigma A T^4$

where $\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8}$ W.m$^{-2}$.K$^{-4}$), $A$ is the surface area of the star, and $T$ is its temperature. Since we are only interested in the power radiated in a specific spectral range, we need to use the Planck radiation law:

$B_\lambda(T) = \frac{2hc^2}{\lambda^5} \frac{1}{\exp(hc/\lambda k_BT) - 1}$

to calculate the spectral radiance of the star at the wavelengths of interest. Here $h$ is the Planck constant, $c$ is the speed of light, $\lambda$ is the wavelength, and $k_B$ is the Boltzmann constant. We can integrate this expression over the wavelength range to obtain the total power radiated per unit area in that range:

$P_\text{range} = \int_{599\text{ nm}}^{609\text{ nm}} B_\lambda(T) d\lambda$

To perform the integration, we can use numerical methods or approximations such as the trapezoidal rule or Simpson's rule. Here, we will use the trapezoidal rule:

$P_\text{range} \approx \frac{\Delta\lambda}{2} \left[B_{599\text{ nm}}(T) + B_{609\text{ nm}}(T)\right]$

where $\Delta\lambda = 10$ nm is the width of the spectral range. Substituting the expression for $B_\lambda(T)$ and evaluating the integral, we get:

$P_\text{range} \approx 2.93 \times 10^4$ W.m$^{-2}$

Finally, we can convert this to MW.m$^{-2}$:

$P_\text{range} \approx 0.0293$ MW.m$^{-2}$

Therefore, the power per unit area radiated by the star in the spectral range from 599 nm to 609 nm is 0.03 MW.m$^{-2}$, accurate to 2 decimal places.