Given that \(E\) (in W\(\cdot\)m\(^{-2}\)) is the total hemispherical emissive power of a surface maintained at a certain temperature, which of the following statements is/are CORRECT?
Show Hint
Spectral quantities (\(E_\lambda\), \(I_\lambda\)) depend on \(\lambda\); total quantities (\(E\)) are integrals over \(\lambda\) and direction.
For a blackbody, \(E=\sigma T^4\); for a gray body, \(E=\varepsilon \sigma T^4\).
- \(E\) does not depend on the direction of the emission.
- \(E\) depends on the viewfactor.
- \(E\) depends on the wavelength of the emission.
- \(E\) does not depend on the frequency of the emission.
The Correct Option is A, D
Solution and Explanation
Step 1: The total hemispherical emissive power is defined as \[ E = \int_{\Omega = 2\pi} \int_{0}^{\infty} I_\lambda(\theta,\phi)\,\cos\theta \, d\Omega \, d\lambda, \] or, for a gray surface, \[ E = \varepsilon \sigma T^{4}. \] It is integrated over all directions and all wavelengths.
Step 2: Since the integration is over the full hemisphere, \(E\) is **not a function of the direction**. Therefore, statement (A) is true. Viewfactor pertains to radiative exchange between different surfaces, not to a single surface’s emissive power. Hence, statement (B) is false.
Step 3: Because \(E\) is integrated over wavelength/frequency, the **total hemispherical** \(E\) is not a function of wavelength or frequency variables themselves (though it depends on the spectral properties of the surface through the integral). Hence, statement (C) is false and statement (D) is true.
Final Answer: \[ \boxed{\;\; (A)\ \text{and}\ (D)\ \text{are true} \;} \]
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