QJEEM15Ph45ba: JEE Main 2015 Physics - Class 12 JEEM15Ph45ba Solution: A red emits light at watt uniformly around it. The amplitude of …

Intensity of light at a distance r, $I = {P \over {4\pi {r^2}}}$ [P = power]

Again, if the amplitude of the electric field is E₀

then $I = {1 \over 2}c{ \in _0}E_0^2$

$\therefore$ ${P \over {4\pi {r^2}}} = {1 \over 2}c{ \in _0}E_0^2$

or, ${E_0} = \sqrt {{P \over {2\pi c{ \in _0}{r^2}}}}$

$= \sqrt {{{0.1} \over {2 \times 3.14 \times (3 \times {{10}^8}) \times (8.85 \times {{10}^{ - 12}}) \times {1^2}}}}$

= 2.45 V/m