A monochromatic plane wave of wavelength - -600- m is propagating in the direction as shown in the figure below- Ei-Er- and Et denote incident reflected- and transmitted electric field vectors associated with the wave The angle of incidence - i and the expression for E are

Ans : (c) E_i propagates in ZX plane nbsp; E_i directed in ZX plane It is nbsp; a Perpendicuclar plane of incidence polarized wave. Using Snells law, s ...

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Reflection of Plane Wave at Oblique Incidence (Part 3)

A monochromat...

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A monochromatic plane wave of wavelength $λ = 600 μ m$ is propagating in the direction as shown in the figure below. $_{i},_{r}, a n d_{t}$ denote incident reflected, and transmitted electric field vectors associated with the wave

The angle of incidence $θ_{i}$ and the expression for $\to E$ are

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Solution

Ans : (c)

$E_{i}$ propagates in ZX-plane

$E_{i}$ directed in ZX-plane

It is a Perpendicuclar-plane of incidence polarized wave.

Using Snells law,

$\frac{s i n θ_{i}}{s i n {19.2}^{\circ}} = \sqrt{\frac{4.5 ϵ_{0}}{ϵ_{0}}} => θ_{i} = 45^{\circ}$

$β_{x} = β_{z} = β c o s 45^{\circ} = \frac{β}{\sqrt{2}}$

(for propagation direction)

For field direction = $\frac{{^a}_{x} - {^a}_{z}}{\sqrt{2}} (45^{\circ})$

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A monochromatic plane wave of wavelength λ=600μm is propagating in the direction as shown in the figure below. −→Ei,−→Er,and−→Et denote incident reflected, and transmitted electric field vectors associated with the wave The angle of incidence θi and the expression for →E are

Ans : (c) Ei propagates in ZX-plane Ei directed in ZX-plane It is a Perpendicuclar-plane of incidence polarized wave. Using Snells law, sinθisin19.2∘=√4.5ϵ0ϵ0=>θi=45∘ βx=βz=βcos45∘=β√2 (for propagation direction) For field direction = ^ax−^az√2(45∘)