Practice portal › Electromagnetic Waves › Intensity, Power and Amplitude

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

Asked in JEE Main 2015 · Point source, efficiency and power on an area

(1) 5.48 V/m
(2) 7.75 V/m
(3) 1.73 V/m
(4) 2.45 V/m

Answer: (4) 2.45 V/m

Step-by-step solution

I = P/4π r² = 0.1/(4π × 1) = 7.96 × 10⁻³ W m⁻²

I = 1/2ε₀cE₀² ⇒ E₀ = √2I/ε₀c = (2 × 7.96 × 10⁻³ / 2.655 × 10⁻³)

E₀ = √5.99 = 2.45 V/m

Why the other options are wrong

(1) 5.48 V/m is √2 × … from using I = P/4π r² without the 1/2, then doubling again
(2) 7.75 V/m is √10 × 2.45 — a decade slipped into the 0.1 W
(3) 1.73 V/m = 2.45/√2, the rms field; the question asks for the amplitude

More Intensity, Power and Amplitude questions

All Intensity, Power and Amplitude questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer