Question 15

A point source of $$100$$ W emits light with $$5\%$$ efficiency. At a distance of $$5$$ m from the source, the intensity produced by the electric field component is:

$$P_{\text{light}} = P \times \eta\% = 100 \times \frac{5}{100} = 5\text{ W}$$

$$I_{\text{total}} = \frac{P_{\text{light}}}{4\pi r^2}$$

$$I_{\text{total}} = \frac{5}{4\pi (5)^2} = \frac{5}{100\pi} = \frac{1}{20\pi}\text{ W/m}^2$$

$$I_E = I_B = \frac{1}{2}I_{\text{total}}$$

$$I_E = \frac{1}{2} \times \frac{1}{20\pi} = \frac{1}{40\pi}\text{ W/m}^2$$

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