A plane electromagnetic wave travelling in vacuum is incident normally on a non-magnetic, non-absorbing medium of refractive index n. The incident (Ei), reflected (Er) and transmitted (Et) electric fields are given as
Ei = E exp[i(kz - ωt)], Er = E0r exp[i(krz - ωt)], Et = E0t exp[i(ktz - ωt)]
If E = 2 V/m and n = 1.5, then the application of of appropriate boundary condition what is magnitude of Eor,Eot?

Question

Answer

Incident Electric Field (Ei)
The incident electric field is given by:
Ei = E exp[i(kz - ωt)]
Where:
- E = 2 V/m (magnitude of the incident electric field)
- k = wave vector in vacuum

Refractive Index and Wave Vector
For a medium with refractive index n = 1.5, the wave vector in the medium (kt) is related to the wave vector in vacuum (k) by:
kt = k/n
Where:
- k = ω/c (c = speed of light in vacuum)

Boundary Conditions for Electric Field
At the boundary between two media, the boundary conditions dictate that the tangential components of the electric field must be continuous. Therefore:
Ei + Er = Et
Where:
- Er = E0r exp[i(krz - ωt)] (reflected electric field)
- Et = E0t exp[i(ktz - ωt)] (transmitted electric field)

Magnitude of Reflected and Transmitted Fields
Given the refractive index and the incident electric field, we can express the reflected and transmitted fields using Fresnel's equations. For normal incidence, the reflection coefficient (R) and transmission coefficient (T) are given by:
- R = |E0r/E| = |(n - 1)/(n + 1)|²
- T = |E0t/E| = 1 - R
Calculating R for n = 1.5:
R = |(1.5 - 1)/(1.5 + 1)|² = |0.5/2.5|² = 0.04
Thus:
E0r = E * √R = 2 * √0.04 = 0.4 V/m
Now, to find E0t:
T = 1 - R = 0.96, hence:
E0t = E * √T = 2 * √0.96 ≈ 1.95 V/m

Final Results
- Magnitude of the reflected electric field (E0r): **0.4 V/m**
- Magnitude of the transmitted electric field (E0t): **1.95 V/m**

Top Courses for Physics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams