JEE Main · 2023 · 30 Jan 2023 · Shift 2 · EM Waves

A point source of 100 W emits light with

5%

efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is :

(A) $\frac{1}{2 π} \frac{W}{m ^{2}}$
(B) $\frac{1}{40 π} \frac{W}{m ^{2}}$ ✓ Correct
(C) $\frac{1}{10 π} \frac{W}{m ^{2}}$
(D) $\frac{1}{20 π} \frac{W}{m ^{2}}$

Answer: (B)

Solution

Only

5%

of the

100

W emerges as light, giving a radiated power of

5

W spread over a sphere of radius

5

m, so the total intensity is

I = \frac{P}{4 π r ^{2}} = \frac{5}{4 π \times 25} = \frac{1}{20 π}

^{- 2}

. In an electromagnetic wave the energy is shared equally between the electric and magnetic fields, so the share carried by the electric field component is half of this,

\frac{1}{2} \times \frac{1}{20 π} = \frac{1}{40 π}

^{- 2}

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option

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