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Electromagnetic Waves question

2021 · 24 Feb · Shift 2 · Q62
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  5. /2021 · 24 Feb · Shift 2 · Q62

Electromagnetic Waves question

2021 · 24 Feb · Shift 2 · Q62

JEE MainPhysicsElectromagnetic WavesNumerical+4 / −1
An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be ‾×\underline{\hspace{2cm}}\times​× 10 −-− 2 cm.
Numerical answer
View written solutionFree

Correct answer: 667

  1. Given data
  • Frequency of electromagnetic wave: f=3 GHz=3×109 Hzf = 3\,\text{GHz} = 3\times 10^9\,\text{Hz}f=3GHz=3×109Hz
  • Relative permittivity of dielectric medium: εr=2.25\varepsilon_r = 2.25εr​=2.25

For a non-magnetic dielectric medium, μr=1\mu_r = 1μr​=1 so the refractive index is n=εr=2.25=1.5n = \sqrt{\varepsilon_r} = \sqrt{2.25} = 1.5n=εr​​=2.25​=1.5

  1. Speed of wave in the medium

Speed in medium is v=cn=3×1081.5=2×108 m/sv = \frac{c}{n} = \frac{3\times 10^8}{1.5} = 2\times 10^8\,\text{m/s}v=nc​=1.53×108​=2×108m/s

  1. Wavelength in the medium

Using λ=vf\lambda = \frac{v}{f}λ=fv​ we get λ=2×1083×109\lambda = \frac{2\times 10^8}{3\times 10^9}λ=3×1092×108​ λ=23×10−1 m\lambda = \frac{2}{3}\times 10^{-1}\,\text{m}λ=32​×10−1m λ=0.0667 m\lambda = 0.0667\,\text{m}λ=0.0667m

  1. Convert into cm

Since 1 m=100 cm1\,\text{m} = 100\,\text{cm}1m=100cm

λ=0.0667×100=6.67 cm\lambda = 0.0667\times 100 = 6.67\,\text{cm}λ=0.0667×100=6.67cm

Now write it in the required form: 6.67 cm=667×10−2 cm6.67\,\text{cm} = 667\times 10^{-2}\,\text{cm}6.67cm=667×10−2cm

So the blank is: 667\boxed{667}667​

  1. Comparison with stored answer

Stored correct answer = 667

Our derived answer = 667

They match.

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