© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the power and the distance
We are given that the power of the red LED is $0.1\,\text{W}$, and it emits light uniformly in all directions. The distance at which we want to find the electric field amplitude is $1\,\text{m}$ from the LED.
Step 2: Write down the formula for intensity from a point source
Because the LED emits uniformly in all directions, the intensity $I$ of the light at distance $r$ is:
$I = \frac{P}{4\pi r^2},$
where $P$ is the power emitted and $r$ is the distance from the source.
Step 3: Relate intensity to the amplitude of the electric field
For an electromagnetic wave in free space, the intensity $I$ is related to the electric field amplitude $E_0$ by the expression:
$I = \frac{1}{2}\,c\,\epsilon_0\,E_0^2,
where $c$ is the speed of light in vacuum and $\epsilon_0$ is the permittivity of free space.
Step 4: Express $E_0$ in terms of power and constants
Substituting $I = \frac{P}{4 \pi r^2}$ into $I = \frac{1}{2} c \epsilon_0 E_0^2$ and then solving for $E_0$, we get:
$\frac{P}{4\pi r^2} = \frac{1}{2}c\,\epsilon_0\,E_0^2.
$
So,
$
E_0 = \sqrt{\frac{2P}{4\pi\,r^2\,c\,\epsilon_0}}
= \sqrt{\frac{P}{2\pi\,r^2\,c\,\epsilon_0}}.
$
Step 5: Substitute the numerical values
Use the following constant values:
$c = 3 \times 10^8\,\text{m/s}$,
$\epsilon_0 = 8.85 \times 10^{-12}\,\text{F/m}$,
$\pi \approx 3.14$,
$P = 0.1\,\text{W}$,
$r = 1\,\text{m}$.
So,
$
E_0 = \sqrt{\frac{0.1}{2 \pi \times 3 \times 10^8 \times 8.85 \times 10^{-12} \times 1^2}}.
$
Step 6: Calculate and conclude the result
Performing the calculation gives:
$
E_0 \approx 2.45\,\text{V/m}.
$
Hence, the amplitude of the electric field of the emitted light at a distance of $1\,\text{m}$ is approximately $2.45\,\text{V/m}$.
