JEE Main 2019 — Electromagnetic Waves Question with Solution
From: JEE Main 2019 (Online) 11th January Evening Slot
Question
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 0 = 9 10–12 SI units, Speed of light c = 3 108 m/s]
[Given permittivity of space 0 = 9 10–12 SI units, Speed of light c = 3 108 m/s]
Choose an option
Show full solutionCorrect option: C
Correct answer
C1.4 kV/m
Step-by-step explanation
Intensity of EM wave is given by
kv/m
kv/m
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This is a previous-year question from JEE Main 2019, covering the Electromagnetic Waves chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.