Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A $30 \mathrm{~mW}$ laser beam has a cross-sectional area of $15 \mathrm{~mm}^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by $\left[\begin{array}{l}\text { Permittivity of space, } \varepsilon_0=9 \times 10^{-12} \\ \text { Speed of light, } c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\end{array}\right]$
PhysicsElectromagnetic WavesCOMEDKCOMEDK 2022
Options:
  • A $1.22 \mathrm{kV} / \mathrm{m}$
  • B $12 \mathrm{kV} / \mathrm{m}$
  • C $10 \mathrm{kV} / \mathrm{m}$
  • D $201 \mathrm{kV} / \mathrm{m}$
Solution:
1599 Upvotes Verified Answer
The correct answer is: $1.22 \mathrm{kV} / \mathrm{m}$
Given, power of laser beam,
$$
P=30 \mathrm{~mW}=30 \times 10^{-3} \mathrm{~W}
$$
Area of cross section, $A=15 \mathrm{~mm}^2=15 \times 10^{-6} \mathrm{~m}^2$
Permittivity of free space, $\varepsilon_0=9 \times 10^{-12}$ SI unit
Speed of light, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Intensity of electromagnetic wave is given by
$$
I=\frac{1}{2} n c \varepsilon_0 E^2
$$
where, $n$ is refractive index for air, $n=1$
$$
\begin{aligned}
I & =\frac{1}{2} c \cdot \varepsilon_0 E^2 \\
I & =\frac{P}{A}
\end{aligned}
$$


From Eqs. (i) and (ii), we get
$$
\begin{aligned}
\frac{1}{2} c \varepsilon_0 E^2 & =\frac{P}{A} \\
E^2 & =\frac{2 P}{A c \varepsilon_0} \\
E & =\sqrt{\frac{2 \times 30 \times 10^{-3}}{15 \times 10^{-6} \times 3 \times 10^8 \times 9 \times 10^{-12}}} \\
& =1.22 \times 10^3 \mathrm{~V} / \mathrm{m}=1.22 \mathrm{kV} / \mathrm{m}
\end{aligned}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.