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JEE Main 2021 (17 Mar, Shift 2) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2021 (17 Mar, Shift 2)

Q22JEE Main 2021 · 17 MAR · SHIFT 2Medium

Energy Density & Intensity

The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60W at the same distance is $\sqrt{\frac{x}{5}}$ E. Where the value of x = ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

$I = \dfrac{P}{4\pi r^2} = \tfrac{1}{2}\varepsilon_0 c E_0^2$ , so at the same $r$ , $E_0 \propto \sqrt{P}$
$\dfrac{E'}{E} = \sqrt{\dfrac{60}{100}} = \sqrt{\dfrac{3}{5}}$
So $x = 3$ .

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