An electric bulb of power 30 π W has an efficiency of 10 % and it can be assumed to behave like a point source of light. At a distance of 3 m from the bulb, the peak value of the electric field in the

Question

An electric bulb of power 30 π W has an efficiency of 10 % and it can be assumed to behave like a point source of light. At a distance of 3 m from the bulb, the peak value of the electric field in the light produced by the bulb is [Take ε 0 ≈ 9 × 10 - 12 C 2 N - 1 m - 2 ]

TestHub · Accepted Answer

The bulb's input power is 30pi W and efficiency is 0.10. Hence its light output power is $P_{out} = η P_{in} = 0.10 (30 π) = 3 π W, I = \frac{P _{out}}{4 π r ^{2}} = \frac{3 π}{4 π ( 3 ) ^{2}} = \frac{1}{12} W m^{- 2}$

For an electromagnetic wave, $I = \frac{1}{2} c ϵ_{0} E_{0}^{2}, E_{0} = \frac{2 I}{c ϵ _{0}} = \frac{1}{6 ( 3 \times 1 0 ^{8} ) ( 9 \times 1 0 ^{- 12} )} = \frac{100}{9 2} V m^{- 1}$

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JEE Advanced 2027 (Class 12)