Question

A $\mathbf{100}\mathbf{}\mathit{W}$ light bulb is placed in a fixture with a reflector that makes a spot of radius $\mathbf{20}\mathbf{}\mathit{c}\mathit{m}$. Calculate approximately the amplitude of the radiative electric field in the spot.

Short answer

The approximate the amplitude of the radiative electric field in the spot is $774\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.

Step-by-step solution

Identification of given data

The power of light bulb,$P=100\text{W}$

The radius of the spot,$r=20\text{cm}$

Determine the formulas to calculate the approximately the amplitude of the radiative electric field in the spot.

The intensity is defined as the ratio of the power and the area.

The expression to calculate the intensity in terms of power and area is given as follows.

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$ …(i)

Here, A is the area.

The expression to calculate the intensity of the electromagnetic wave is given as follows.

$\mathit{I}\mathbf{=}\frac{{\mathbf{E}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{c}}$ …(ii)

Here,${\mathit{\mu }}_{\mathbf{0}}$ is the permeability of the vacuum, c is the speed of the light and E is the radiative electric field.

Determining the approximately the amplitude of the radiative electric field in the spot.

Derive the expression for the radiative electric field from the equation (ii).

$$\begin{gathered} E^2=2{\mu }_{0}cI \\ E=\sqrt{2{\mu }_{0}cI} \end{gathered}$$

Substitute $\raisebox{1ex}{$P$}\!\left/ \!\raisebox{-1ex}{$A$}\right.forI$into above equation.

$E=\sqrt{\frac{2{\mu }_{0}cP}{A}}$

Substitute $\pi r^{2}forA$into above equation.

$E=\sqrt{\frac{2{\mu }_{0}cP}{\pi r^2}}$

Substitute $100\text{WforP,}4\pi \times {10}^{-7}T.\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$Afor{\mu }_0,20cmforr,and3\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}$}\right.forc$, into above equation.

$$\begin{gathered} E=\sqrt{\frac{2\times \left({\displaystyle 4\pi \times {10}^{-7}T.\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}\right)\times \left({\displaystyle 3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\right){\displaystyle \times }{\displaystyle 100}{\displaystyle }{\displaystyle \text{W}}}{{\displaystyle \pi \times {\left(0.20m\right)}^2}}} \\ =\sqrt{\frac{{\displaystyle 7.53\times {10}^{4}T.m^2\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$A.s$}\right.}}{{\displaystyle 0.1256{\text{m}}^2}}} \\ =\sqrt{59.92\times {10}^{4}T.\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$A.s$}\right.} \\ =774\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right. \end{gathered}$$

Hence the approximate the amplitude of the radiative electric field in the spot is $774\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.