A lamp emits monochromatic green light uniformly in all directions. lamp is 3% efficient in converti

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8.Electromagnetic waves

hard

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.......$V/m$

$1.34$

$2.68$

$4.02$

$5.36 $

(JEE MAIN-2014)

Solution

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)}}} $

$=\sqrt{\frac{6}{25} \times 30}=\sqrt{7.2}$

$\therefore \mathrm{E}_{0}=2.68 \mathrm{V} / \mathrm{m}$

Standard 12

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Solution

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A flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave

$E_x=36\,sin\,(1.20 \times 10^7z -3.6 \times 10^{15}\,t)\,V/m$

The average intensity of the beam will be…..$W/m^2$