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A plane electromagnetic wave of frequency (1.5 ,\text{Hz}) travels in a dielectric material with relative permittivity (=4.5) and relative permeability (=1). If the incident amplitude in air is (2000,\text{V m}^{-1}) and the wave strikes the dielectric at normal incidence, what is the magnitude of the electric field amplitude inside the dielectric?

hard MCQ 3 marks Class 12 · Physics · Electromagnetic Waves
  1. (1.28 ,\text{V m}^{-1})
  2. (1.58 ,\text{V m}^{-1})
  3. (1.00 ,\text{V m}^{-1})
  4. (0.64 ,\text{V m}^{-1})
Answer
(1.28 ,\text{V m}^{-1})

The correct answer is (1.28 ,\text{V m}^{-1}) — Step 5: Multiply by (2000) to get (=0.6405\times2000=1281,\text{V m}^{-1}).

Step-by-step solution

  1. Step 1: Compute the refractive index of the dielectric: (n=sqrt{}=sqrt{4.5}=2.1213).
  2. Step 2: For normal incidence, the transmitted electric field amplitude is given by (= \frac{}{}), where (=1) (air) and (=n).
  3. Step 3: Substitute the values: (= \frac{2\times1}{1+2.1213}\times2000 = \frac{2}{3.1213}\times2000).
  4. Step 4: Evaluate the fraction: (\frac{2}{3.1213}=0.6405).
  5. Step 5: Multiply by (2000) to get (=0.6405\times2000=1281,\text{V m}^{-1}).
Common mistake: Using the reflection coefficient instead of the transmission coefficient

Concept tested: Transmission coefficient for normal incidence

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