Physics Circle home Student Login

Home›Chapters›Electromagnetic Waves PYQs›Energy Density & Intensity›Question 30

JEE Main 2019 (11 Jan, Shift 2) · Energy Density & Intensity

Energy Density & Intensity question from JEE Main 2019 (11 Jan, Shift 2)

Q30JEE Main 2019 · 11 JAN · SHIFT 2Medium

Energy Density & Intensity

A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by:

[Given permittivity of space ε\varepsilon0 = 9 ×\times 10–12 SI units, Speed of light c = 3 ×\times 108 m/s]

Show answer and solution

Answer: (c)

  1. I=PA=27×10−310×10−6=2700I = \dfrac{P}{A} = \dfrac{27 \times 10^{-3}}{10 \times 10^{-6}} = 2700 W/m2^2
  2. E0=2Iε0c=54009×10−12×3×108=2×106E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = \sqrt{\dfrac{5400}{9 \times 10^{-12} \times 3 \times 10^{8}}} = \sqrt{2 \times 10^{6}}
  3. E0≈1414E_0 \approx 1414 V/m ≈1.4\approx 1.4 kV/m
  4. So the answer is (c) 1.4 kV/m.
Report a mistake

More questions on Energy Density & Intensity