Question

A lamp emits monochromatic green light uniformly in all directions. The lamp is $3$% efficient in converting electrical power to electromagnetic waves and consumes $100W$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $5m$ from the lamp will be nearly
$\begin{align}
  & A)1.34V{{m}^{-1}} \\
 & B)2.68V{{m}^{-1}} \\
 & C)4.02V{{m}^{-1}} \\
 & D)5.36V{{m}^{-1}} \\
\end{align}$

Answer 1

Hint: Intensity of electromagnetic radiation at a particular point is equal to the ratio of power of electromagnetic radiation at that point to the area of the sphere whose radius is equal to the distance of the considered point from the lamp. Half of electromagnetic intensity is due to electric field while the other half of electromagnetic intensity is due to magnetic field. Intensity of electromagnetic radiation at a point is also proportional to the square of electric field associated with the electromagnetic radiation, at that point.
Formula used:
$I=\dfrac{Power}{Area}$
$I=\dfrac{1}{2}{{\varepsilon }_{0}}c{{E}_{0}}^{2}$

Complete answer:
We know that electromagnetic intensity at a particular point is equal to the ratio of power of radiation at that point to the area of the sphere whose centre is considered to be the position of the considered point. Clearly, the radius of the sphere is equal to the distance of the considered point from the source of electromagnetic radiation. Mathematically, electromagnetic intensity is given by
$I=\dfrac{Power}{Area}=\dfrac{P}{4\pi {{R}^{2}}}$
where
$I$ denotes the electromagnetic intensity at a particular point
$Power=P$ is the power of electromagnetic radiation at that point
$Area=4\pi {{R}^{2}}$ is the area of sphere whose radius is equal to the distance of the considered point from the source of electromagnetic radiation
Let this be equation 1.
Clearly, if $d$ denotes the distance of the point from the source of electromagnetic radiation, then, equation 1 becomes
$I=\dfrac{P}{4\pi {{d}^{2}}}$
Let this be equation 2.
Electromagnetic intensity at a point is the sum of electromagnetic intensity due to electric field as well as electromagnetic intensity due to magnetic field at that point. This can be expressed as
$I={{I}_{E}}+{{I}_{M}}=\dfrac{I}{2}+\dfrac{I}{2}$
where
$I$ is the electromagnetic intensity at a point
${{I}_{E}}=\dfrac{I}{2}$ is the electromagnetic intensity due to the electric field at that point
${{I}_{M}}=\dfrac{I}{2}$ is the electromagnetic intensity due to the magnetic field at that point
Let this be equation 3.
It is clear from equation 3 that intensities due to both electric field as well as magnetic field are equal and half the total electromagnetic intensity at a given point.
Now, we also know that electromagnetic intensity at a point is proportional to the square of amplitude of the electric field at that point. Thus, electromagnetic intensity can also be expressed as
$I=\dfrac{1}{2}{{\varepsilon }_{0}}c{{E}^{2}}$
where
$I$ is the electromagnetic intensity at a point
$E$ is the amplitude of electric field at that point
${{\varepsilon }_{0}}$ is the permittivity of free space
$c$ is the speed of light in free space
Let this be equation 4.
Considering the intensity due to electric field and thus, combining equation 3 and equation 4, we have
 ${{I}_{E}}=\dfrac{I}{2}=\dfrac{1}{4}{{\varepsilon }_{0}}c{{E}^{2}}$
where
${{I}_{E}}=\dfrac{I}{2}$ is the electromagnetic intensity due to the electric field at a point
Let this be equation 5.
Coming to our question, we are provided with a lamp, which emits monochromatic green light uniformly in all directions. The lamp is said to be $3$% efficient in converting electrical power to electromagnetic waves and consumes $100W$ of power. We are required to determine the amplitude of the electric field associated with the electromagnetic radiation at a distance of $5m$ from the lamp.
Clearly, power of the lamp at a distance $d=5m$ from the lamp is given by
$P=\dfrac{3}{100}\times 100W=3W$
Let this be equation 6.
Also, if we consider a sphere whose radius is equal to the distance of the point from the lamp $(d=5m)$, then, area of the sphere is given by
$Area=4\pi {{d}^{2}}=4\pi {{(5m)}^{2}}=314{{m}^{2}}$
Let this be equation 7.
Substituting equation 6 and equation 7 in equation 2, we have
$I=\dfrac{P}{4\pi {{d}^{2}}}=\dfrac{3W}{314{{m}^{2}}}=9.5\times {{10}^{-3}}W{{m}^{-1}}$
where
$I$ is the intensity of monochromatic green light
Let this be equation 8.
Using equation 5, intensity of monochromatic light due to electric field at a distance $d=5m$ from the lamp is given by
\[{{I}_{E}}=\dfrac{I}{2}=\dfrac{1}{4}{{\varepsilon }_{0}}c{{E}^{2}}\Rightarrow {{I}_{E}}=\dfrac{9.5\times {{10}^{-3}}}{2}=\dfrac{1}{4}{{\varepsilon }_{0}}c{{E}^{2}}\]
where
${{I}_{E}}=\dfrac{I}{2}$ is the intensity of monochromatic green light due to the electric field at a distance $d=5m$ from the lamp
$I=9.5\times {{10}^{-3}}W{{m}^{-1}}$ is the intensity of monochromatic green light, from equation 8
$E$ is the amplitude of electric field at a distance $d=5m$ from the lamp
${{\varepsilon }_{0}}$ is the permittivity of free space
$c$ is the speed of light in free space
Let this be equation 9.
Simplifying equation 9, we have
$\dfrac{9.5\times {{10}^{-3}}}{2}=\dfrac{1}{4}{{\varepsilon }_{0}}c{{E}^{2}}\Rightarrow 4.75\times {{10}^{-3}}=\dfrac{1}{4}\times (8.85\times {{10}^{-12}})\times (3\times {{10}^{8}}){{E}^{2}}$
On further simplification, we have
$4.75\times {{10}^{-3}}=6.6375\times {{10}^{-4}}\times {{E}^{2}}\Rightarrow {{E}^{2}}=\dfrac{4.75\times {{10}^{-3}}}{6.6375\times {{10}^{-4}}}\approx 7.2\Rightarrow E=\sqrt{7.2}=2.68V{{m}^{-1}}$
Let this be equation 10.
Therefore, from equation 10, we can conclude that the amplitude of the electric field associated with the electromagnetic radiation at a distance of $5m$ from the lamp is nearly $2.68V{{m}^{-1}}$.

Hence, the correct answer is option $B$.

Note:
Students need to be thorough with the values of speed of light in free space as well as permittivity of free space to solve this problem easily. For reference, they are given as follows:
$c=3\times {{10}^{8}}m{{s}^{-1}}$, is the speed of light in free space
${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}F{{m}^{-1}}$ is the permittivity of free space
It is also important to notice from equation 3 that intensities due to both electric field as well as magnetic field at a given point are the same, which is nothing but half the value of electromagnetic intensity at that given point.