Electromagnetic Waves

JEE Main level, single correct, medium

126 of 165

Question 126 of 165

A lamp emits monochromatic green light uniformly in all directions. The lamp is 3%3 \% efficient in converting electrical power to electromagnetic waves and consumes 100 W100 \mathrm{~W} of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m5 \mathrm{~m} from the lamp will be nearly:

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Answer: (B) 2.68 V/m2.68 \mathrm{~V} / \mathrm{m}

Wavelength of monochromatic green light
=5.5×10−5 cm Intensity I = Power  Area =100×(3/100)4π(5)2=3100πWm−2\begin{aligned} &=5.5 \times 10^{-5} \mathrm{~cm} \\ &\text { Intensity I }=\frac{\text { Power }}{\text { Area }} \\ &=\frac{100 \times(3 / 100)}{4 \pi(5)^2} \\ &=\frac{3}{100 \pi} \mathrm{Wm}^{-2} \end{aligned}
Now, half of this intensity (I) belongs to electric field and half of that to magnetic field, therefore,
I2=14ε0E02C or E0=2Iε0C\begin{aligned} &\frac{\mathrm{I}}{2}=\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2 \mathrm{C} \\ &\text { or } \mathrm{E}_0=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_0 \mathrm{C}}} \end{aligned}

=2×(3100π)(14π×9×109)×(3×108=625×30=7.2∴E0=2.68 V/m\begin{aligned} &=\sqrt{\frac{2 \times\left(\frac{3}{100} \pi\right)}{\left(\frac{1}{4 \pi \times 9 \times 10^9}\right) \times\left(3 \times 10^8\right.}} \\ &=\sqrt{\frac{6}{25} \times 30}=\sqrt{7.2} \\ &\therefore \mathrm{E}_0=2.68 \mathrm{~V} / \mathrm{m} \end{aligned}

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