JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2014
- Hard
- MCQ
A lamp emits monochromatic green light uniformly in all directions. The lamp is \(3\%\) efficient in converting electrical power to electromagnetic waves and consumes \(100\,W\) of power . The amplitude of the electric field associated with the electromagnetic radiation at a distance of \(5\,m\) from the lamp will be nearly.......\(V/m\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2.68\)
Step-by-step Solution
Detailed explanation
Wavelength of monochromatic green light \(=5.5 \times 10^{-5} \mathrm{cm}\)
Intensity \(I\,=\,\frac{\text { Power }}{\text { Area }}\)
\( = \frac{{100 \times (3/100)}}{{4\pi {{(5)}^2}}}\)
\(=\frac{3}{100 \pi} \mathrm{Wm}^{-2}\)
Now, half of this intensity ( \(\mathrm{I}\) ) belongs to electric field and half of that to magnetic field, therefore,
\(\frac{1}{2}=\frac{1}{4} \varepsilon_{0} E_{0}^{2} C\)
or \(\mathrm{E}_{0}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{C}}}\)
\( = \sqrt {\frac{{2 \times \left( {\frac{3}{{100}}\pi } \right)}}{{\left( {\frac{1}{{4\pi \times 9 \times {{10}^9}}}} \right) \times \left( {3 \times {{10}^8}} \right)}}} \)
\(=\sqrt{\frac{6}{25} \times 30}=\sqrt{7.2}\)
\(\therefore \mathrm{E}_{0}=2.68 \mathrm{V} / \mathrm{m}\)
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