Applications of Electromagnetic Waves | Electricity & Magnetism - Physics PDF Download

Reflection and Refraction at Dielectric Interface

Normal Incidence

Suppose xy plane forms the boundary between two linear media. A plane wave of frequency ω , traveling in the z-direction and polarized in the x direction, approaches the interface from the left then

Incident Wave 

Reflected Wave

Transmitted Wave

At z = 0 , the combined field on the left  must join the fields on the right  in accordance with the boundary conditions

(i)
(ii)
(iii)
(iv)  
In this case there are no electric component perpendicular to the surface, so (i) & (ii) are trivial. However (iii) gives

While (iv) gives,

where

Solving above two equations we get

if  (For non-magnetic medium)

Note: Reflected wave is in phase if v₂ >v₁ or n₂< n₁ and out of phase if v₂ <v₁ or n₂> n₁.

Since Intensity  then the ratio of the reflected intensity to the incident intensity is the Reflection coefficient

The ratio of the transmitted intensity to the incident intensity is the Transmission coefficient

Example 1:  Calculate the reflection coefficient for light at an air-to-dielectric interface
(μ₁ =μ₂= μ₀ , n₁ =1,n₂ = 1.5) at optical frequencyω = 4×10¹⁵ s⁻¹.

Reflection coefficient

Thus only 4% of light is reflected and 96% is transmitted.

Oblique Incidence

In oblique incidence an incoming wave meets the boundary at an arbitrary angle θ_I . Of course, normal incidence is really just a special case of oblique incidence with θ_I = 0 . Suppose that a monochromatic plane wave of  frequency ω , approaches the interface from the left then

Incident Wave

Reflected Wave

Transmitted Wave

All three waves have the same frequencyω . The three wave numbers are related by (ω = kv ) as

The combined field in medium (1),  must join the fields  in medium (2), using the boundary conditions.
(i)  
(ii)
(iii)
(iv)
First Law (Plane of Incidence)
The incident, reflected and transmitted wave vectors form a plane (called the plane of incidence), which also includes normal to the surface.

Second law (Law of Reflection)
The angle of incidence is equal to the angle of reflection i.e.
 θ_I= θ_R
Third Law: (Law of Refraction, or Snell’s law)
 
Fresnel’s Relation (Parallel and Perpendicular Polarization) Case-I: (Polarization in the Plane of Incidence) Applying Boundary conditions, we get Reflected and transmitted amplitudes

 
These are known as Fresnel’s equations.

Notice that transmitted wave is always in phase with the incident one; the reflected wave is either in phase, if   α > β , or 180⁰ out phase if α < β .
The amplitudes of the transmitted and reflected waves depend on the angle of incidence, because α is a function of θ_I :

Brewster’s Angle

At Brewster’s angle (θ_B ) reflected light is completely extinguished whenα = β , or  

For non-magnetic medium
 and hence

Thus at Brewster angle  reflected and transmitted rays are perpendicular to each other. 

Critical Angle

When light enters from denser to rarer medium ( n₁ > n₂ ) then after a critical angle (θ_C ) there is total internal reflection.

Reflection and Transmission Coefficient 

The power per unit are striking the interface is  Thus the incident intensity is

while reflected and transmitted intensities are

Reflection coefficient

Transmission coefficient

⇒ R+T = 1

Case-II: (Polarization Perpendicular to plane of Incidence)

Applying Boundary conditions, we get Reflected and transmitted amplitudes

In this case Brewster’s angle (θ_B ) is not possible i.e reflected light is never completely extinguished (since αβ = 1 is not possible).

Reflection and Transmission coefficient

The power per unit are striking the interface is . Thus the incident intensity is

while reflected and transmitted intensities are

Reflection coefficient

Transmission coefficient

⇒ R+T = 1

Reflection at Conducting Surface (Normal Incidence)

Suppose xy plane forms the boundary between a non-conducting linear medium (1) and a conductor (2). A plane wave of frequencyω , traveling in the z-direction and polarized in the x direction, approaches the interface from the left then

Incident Wave

Reflected Wave

Transmitted Wave

where  where k₂ and κ₂ are real and imaginary part of

At z = 0 , the combined field on the left  must join the fields on the right  in accordance with the boundary conditions
(i) 
(ii) 
(iii) 
(iv)  
In this case there are no electric component perpendicular to the surface, so (i) & (ii) are trivial.
However (iii) gives

While (iv) gives,

where

Solving above two equations we get

Note: 
(i) For a perfect conductor  Thus
 
In this case wave is totally reflected, with a 1800 phase shift.
(ii) For good conductor
 

Reflection Coefficient

Example 2:  Calculate the reflection coefficient for light at an air-to-silver interface  μ₁=μ₂= μ₀ , ε₁ = ε₀, σ = 6 ×10⁷ Ω⁻¹ m− at optical frequency ω = 4×10¹⁵ s⁻¹ .

 
Reflection coefficient

Thus 93% of light is reflected and only 7% is transmitted.