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Electromagnetic Waves question

2013 · Q132
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Electromagnetic Waves question

2013 · Q132

NEETPhysicsElectromagnetic WavesMCQ+4 / −1
An electromagnetic wave of frequency υ=3.0\upsilon = 3.0υ=3.0 MHz passes from vaccum into a dielectric medium with relative permittivity εr{{\varepsilon _r}}εr​ = 4.0. Then
  1. A
    Wavelength is doubled and frequency becomes half.
  2. B
    Wavelength is halved and frequency remains unchanged.
  3. C
    Wavelength and frequency both remain unchanged.
  4. D
    Wavelength is doubled and frequency unchanged.
View written solutionFree

Correct answer: B

Velocity of electromagnetic wave in vacuum

c = 1μ0ε0{1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}μ0​ε0​​1​ = νλ\nu \lambda νλvacuum .....(1)

Velocity of electromagnetic wave in the medium

vmedium = 1μ0μrε0εr{1 \over {\sqrt {{\mu _0}{\mu _r}{\varepsilon _0}{\varepsilon _r}} }}μ0​μr​ε0​εr​​1​ = cμrεr{c \over {\sqrt {{\mu _r}{\varepsilon _r}} }}μr​εr​​c​

For dielectric medium, μ\mu μr = 1

∴\therefore∴ vmedium = cεr{c \over {\sqrt {{\varepsilon _r}} }}εr​​c​ = c4=c2{c \over {\sqrt 4 }} = {c \over 2}4​c​=2c​ ........(2)

Wavelength of the wave in medium

λ\lambda λmedium = vmediumν{{{v_{medium}}} \over \nu }νvmedium​​ = c2ν{c \over {2\nu }}2νc​ = λvaccum2{{{\lambda _{vaccum}}} \over 2}2λvaccum​​

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