Question

A red LED emits light at \[0.1\,{\text{Watt}}\] uniformly around it. The amplitude of the electric field of the light at a distance of \[1\,{\text{m}}\] from the diode is:
A. \[6\,{\text{V/m}}\]
B. \[2.45\,{\text{V/m}}\]
C. \[5.48\,{\text{V/m}}\]
D. \[1.73\,{\text{V/m}}\]

Answer 1

Hint: Use the formulae for the intensities of light at a distance r from the source and in terms of the amplitude of the electric field of the light. Determine the relation for amplitude of the electric field of the light by equating these two formulae.

Formulae used:
The formula for the intensity \[I\] of the light at a distance \[r\] from the source is
\[I = \dfrac{P}{{4\pi {r^2}}}\] …… (1)
Here, \[P\] is the power output of the source.
The intensity \[I\] of the light in terms of the amplitude of electric field \[{E_0}\] is
\[I = \dfrac{1}{2}{\varepsilon _0}E_0^2c\] …… (2)
Here, \[{\varepsilon _0}\] is the permittivity of the medium and \[c\] is the speed of the light.

Complete step by step answer:The red LED is the source of light and emits the light uniformly around it.

The power output of the LED light is \[0.1\,{\text{W}}\].
\[P = 0.1\,{\text{W}}\]

Calculate the amplitude of the electric field of the light at a distance of \[1\,{\text{m}}\]from the diode of the LED.

Equate equations (1) and (2) for the intensity of the light.
\[\dfrac{P}{{4\pi {r^2}}} = \dfrac{1}{2}{\varepsilon _0}E_0^2c\]

Rearrange the above equations for \[{E_0}\].
\[ \Rightarrow {E_0} = \sqrt {\dfrac{{2P}}{{4\pi {\varepsilon _0}{r^2}c}}} \]

Substitute \[0.1\,{\text{W}}\] for \[P\], \[9 \times {10^9}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/}}{{\text{C}}^2}\] for \[\dfrac{1}{{4\pi {\varepsilon _0}}}\], \[1\,{\text{m}}\] for \[r\] and \[3 \times {10^8}\,{\text{m/s}}\] for \[c\] in the above equation.
\[ \Rightarrow {E_0} = \dfrac{{2\left( {0.1\,{\text{W}}} \right)\left( {9 \times {{10}^9}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/}}{{\text{C}}^2}} \right)}}{{{{\left( {1\,{\text{m}}} \right)}^2}\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}\]
\[ \Rightarrow {E_0} = \sqrt 6 \]
\[ \Rightarrow {E_0} = 2.45\,{\text{V/m}}\]

Therefore, the amplitude of the electric field of the light at a distance of \[1\,{\text{m}}\]from the diode is \[2.45\,{\text{V/m}}\].

Hence, the correct option is B.

Note:The value of the constant \[\dfrac{1}{{4\pi {\varepsilon _0}}}\] is already calculated as \[9 \times {10^9}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/}}{{\text{C}}^2}\]. Hence, it is substituted directly in the equation for the amplitude of the electric field of the light.