JEE PYQ: Electromagnetic Waves Question 39
Question 39 - 2019 (11 Jan 2019 Shift 2)
A 27 mW laser beam has a cross-sectional area of 10 mm$^2$. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space $\epsilon_0 = 9 \times 10^{-12}$ SI units, Speed of light $c = 3 \times 10^{8}$ m/s]
(1) 2 kV/m
(2) 0.7 kV/m
(3) 1 kV/m
(4) 1.4 kV/m
Show Answer
Answer: (4)
Solution
$I = \frac{P}{A} = \frac{1}{2}\epsilon_0 E_0^2 c$. $E_0 = \sqrt{2 \times 10^{6}} \approx 1.4$ kV/m.
BETA
