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Question #43744

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The Question

A thermal source at absolute temperature $T$ emits unpolarized radiation. The total emitted power per unit area (intensity) follows a modified Stefan-Boltzmann law $I_0 = e(T)\sigma T^4$ , where $\sigma$ is the Stefan-Boltzmann constant and the emissivity is $e(T) = \alpha T$ ( $\alpha$ is a positive constant). The radiation passes through two ideal polarizers. The first polarizer has its transmission axis vertical, and the second has its transmission axis at an angle $\theta$ to the vertical. The transmitted beam then normally enters a non-linear absorbing medium at $x=0$ . The medium has an intensity-dependent absorption coefficient $\mu(I) = \mu_0 + \beta I$ , where $\mu_0$ and $\beta$ are positive constants and $I$ is the local intensity. Assuming all absorbed radiation is converted into heat, what is the volumetric rate of heat generation $q$ at the entry surface $x=0^+$ of the medium?

Options

A.
$\frac{1}{2} \mu_0 \alpha \sigma T^5 \cos^2\theta + \frac{1}{4} \beta \alpha^2 \sigma^2 T^{10} \cos^4\theta$
B.
$\mu_0 \alpha \sigma T^5 \cos^2\theta + \beta \alpha^2 \sigma^2 T^{10} \cos^4\theta$
C.
$\frac{1}{2} \mu_0 \alpha \sigma T^4 \cos^2\theta + \frac{1}{4} \beta \alpha^2 \sigma^2 T^8 \cos^4\theta$
D.
$\frac{1}{2} \mu_0 \alpha \sigma T^5 \cos^2\theta + \frac{1}{2} \beta \alpha^2 \sigma^2 T^{10} \cos^2\theta$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Malus's Law|Stefan-Boltzmann Power Law|Emissivity and Absorptivity|Medium Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of thermal radiation, wave optics, and non-linear absorption, targeting several common student misconceptions.

The primary trap is treating emissivity as a constant, which leads students to incorrectly apply a $T^4$ dependence instead of the $T^5$ dependence dictated by $e(T) = \alpha T$ .

Secondly, students frequently forget that an ideal polarizer halves the intensity of incident unpolarized thermal radiation, missing the crucial $1/2$ factor before applying Malus's Law.

Finally, in the non-linear medium, students may incorrectly calculate the volumetric heat generation by failing to square the intensity in the non-linear term ( $\beta I^2$ ), or by applying the initial intensity $I_0$ to the absorption coefficient instead of the local modulated intensity $I_2$ .

This requires a rigorous application of the differential form of the Beer-Lambert law coupled with non-linear optics.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Wave Optics - Class 12
Concept	Malus's Law
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free