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JEE Main 2020 (6 Sep, Shift 1) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2020 (6 Sep, Shift 1)

Q27JEE Main 2020 · 6 SEP · SHIFT 1Medium

Energy Density & Intensity

Suppose that intensity of a laser is 315π\frac{315}{\pi} W/m2.

The rms electric field, in units of V/m associated with this source is close to the nearest integer is __________.

ε\varepsilon0 = 8.86 × 10–12 C2 Nm–2; c = 3 × 108 ms–1)

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

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Answer: 194

  1. I=c ε0Erms2⇒Erms=Ic ε0I = c\,\varepsilon_0 E_{rms}^2 \Rightarrow E_{rms} = \sqrt{\dfrac{I}{c\,\varepsilon_0}}
  2. Erms2=315/π3×108×8.86×10−12=100.32.658×10−3≈3.77×104E_{rms}^2 = \dfrac{315/\pi}{3 \times 10^8 \times 8.86 \times 10^{-12}} = \dfrac{100.3}{2.658 \times 10^{-3}} \approx 3.77 \times 10^4
  3. Erms≈194 V/mE_{rms} \approx 194\ \text{V/m}
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