A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is:

We are told that the red LED radiates light uniformly in all directions with a total power of $$P = 0.1 \text{ W}$$. When a source sends out energy uniformly, the power spreads over the surface of an imaginary sphere of radius $$r$$ centred on the source. The area of such a sphere is $$4\pi r^{2}$$. Average intensity (power per unit area) at distance $$r$$ is therefore

$$I = \frac{P}{4\pi r^{2}}.$$

We need the intensity at $$r = 1 \text{ m}$$, so we substitute:

$$I = \frac{0.1}{4\pi\,(1)^2} = \frac{0.1}{4\pi} \text{ W m}^{-2}.$$

Numerically, since $$4\pi \approx 12.566$$,

$$I = \frac{0.1}{12.566} \approx 7.96 \times 10^{-3} \text{ W m}^{-2}.$$

Now we connect intensity to the amplitude of the electric field. For a plane electromagnetic wave in free space, the average intensity is given by the standard relation

$$I = \frac{1}{2}\,c\,\epsilon_{0}\,E_{0}^{2},$$

where $$c = 3.0 \times 10^{8} \text{ m s}^{-1}$$ (speed of light) and $$\epsilon_{0} = 8.85 \times 10^{-12} \text{ F m}^{-1}$$ (permittivity of free space). Here $$E_{0}$$ is the required amplitude of the electric field.

We solve this formula for $$E_{0}$$:

$$E_{0} = \sqrt{\frac{2I}{c\,\epsilon_{0}}}.$$

Substituting the numerical values:

First compute the numerator $$2I = 2 \times 7.96 \times 10^{-3} = 1.592 \times 10^{-2} \text{ W m}^{-2}.$$

Next compute the denominator $$c\,\epsilon_{0} = (3.0 \times 10^{8})\,(8.85 \times 10^{-12}) = 2.655 \times 10^{-3}.$$

Now form the ratio inside the square root:

$$\frac{2I}{c\,\epsilon_{0}} = \frac{1.592 \times 10^{-2}}{2.655 \times 10^{-3}} \approx 5.996.$$

Taking the square root,

$$E_{0} = \sqrt{5.996} \text{ V m}^{-1} \approx 2.45 \text{ V m}^{-1}.$$

Thus the amplitude of the electric field 1 m away from the LED is about $$2.45 \text{ V m}^{-1}$$.

Hence, the correct answer is Option C.