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JEE Main 201412 AprilOnline
Exam
JEE Main
Year
2014
Date
12 April
Mode
Online
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2014 · 12 April · Online · Physics · Electromagnetic Waves

A lamp emits monochromatic green light uniformly in all directions....

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:

Options

  1. A1.34 V/m1.34 \mathrm{~V} / \mathrm{m}
  2. B2.68 V/m2.68 \mathrm{~V} / \mathrm{m}
  3. C4.02 V/m4.02 \mathrm{~V} / \mathrm{m}
  4. D5.36 V/m5.36 \mathrm{~V} / \mathrm{m}

Correct answer

B. 2.68 V/m2.68 \mathrm{~V} / \mathrm{m}

Step-by-step solution

Wavelength of monochromatic green light =5.5×10−5cm Intensity I = Power  Area =100×(3/100)4π(5)2=3100πWm−2\begin{aligned}&=5.5 \times10^{-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, \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}\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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