JEE Main 2015 · 4 Apr · Shift 1 · Physics Q23

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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 :

1.73 V/m

2.45 V/m

5.48 V/m

7.75 V/m

Answer and worked solution

Official answer: B

$I=\frac{0.1}{4\pi(1)^2}=7.96\times10^{-3}\text{ W/m}^2,\qquad E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2\times7.96\times10^{-3}}{3\times10^8\times8.85\times10^{-12}}}\approx2.45\text{ V/m}.$

Topics:Energy and momentum of EM waves

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 23