JEE Main 2015PhysicsElectromagnetic WavesMediumMCQ

JEE Main 2015 — Electromagnetic Waves Question with Solution

JEE Main 2015 (04 Apr)

Question

A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:

Choose an option

Show full solutionCorrect option: C
Correct answer
C2.45 V m-1

Step-by-step explanation



For a point source of power = P, then intensity at a point at a separation x from the source is, I=PowerArea=P4πx2∵ The average intensity of E.M. wave is given by, I=12cε0Eo2⇒E0=2P4πε0c x2

∵14πε0=9×109, P=0.1 W, x=1 m, c=3×108 m s-1

⇒E0=2×0.1×9×1093×108×12=6=2.45 V m-1

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About this question

This is a previous-year question from JEE Main 2015, 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.