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For a black body that emits a radiance of $5\times 10^5 Wm^{-2}$ between the wavelengths $\lambda_1=520nm$ and $\lambda_2=550nm$, find the temperature of the given body.

I think we want to work with the equation:

$$R(\lambda,T)=\frac{8\pi hc}{\lambda^5(e^{\frac{hc}{\lambda k_bT}}-1)} $$

And we should integrate in the interval of wavelengths. The obvious problem is that this function has no antiderivative, so it cannot be evaluated in that interval. This means that an approximation is needed.

My first thought was to approximate it by a square of base $30nm$ and height $R(\lambda_1,T)$ or $R(\lambda_2,t)$ and then maybe take the average. Anyway, this hasn't given good results and I don't know why. I would at least want to know if my idea of the approximation is OK.

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Some heat transfer textbooks will include a table listing values of a relevant "blackbody radiation function." In Incropera & DeWitt 6th Ed the function is labelled as

$$ F_{(0 \rightarrow\lambda)} = \frac{1}{\sigma T^4} \int_0^\lambda \frac{2 h c^2}{\lambda^5 \left(e^{\frac{hc}{\lambda k T}}-1 \right)} \,\text{d}\lambda $$

and is tabulated as a function of $\lambda T$. You could get a more precise answer to this problem by using the function $F$, although with the $\lambda$s known but $T$ unknown you would need to use an iterative approach.

The rectangular approximation you propose should give you a rough estimate. I think that the source of your problem may be units/constants. Your original formula has an extra factor of $4\pi/c$ (relative to the value in my $F$ formula above) which means it is providing a spectral energy density (power per area per wavelength) rather than a spectral radiance (power per area per solid angle per wavelength). The value you are taking as a given may also be mislabeled: if it's a radiance, it should be in power per area per solid angle; based on its units, it should be called an energy density.

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