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Matthew Schwartz
Lecture 15:
Refraction and Reflection
1 Refraction
When we discussed polarization, we saw that when light enters a medium with a different index of refraction, the frequency stays the same but the wavelength changes. Using that the speed of light is v = c n we deduced that^ λ^1 n^1 =^ λ^2 n^2 , so that as the index of refraction goes up, the wave- length goes down. The picture looks like this
Figure 1. Plane wave entering and emerging from a medium with different index of refraction.
Now let us ask what happens when light enters a medium with a different index of refraction at an angle. Since we know the wavelength of light in the two media, we can deduce the effect with pictures. The key is to draw the plane waves as the location of the maximum field values. These crests will be straight lines, but spaced more closely together in the medium with higher index of refraction. For example, if sunlight hits ice (or water), the picture looks like this
Figure 2. Matching wavefronts demonstrates refraction. Orange lines represent the crests of waves (or the maximum amplitude
The bending of light when the index of refraction changes is called refraction. To relate the angles θ 1 (the angle of incidence ) to θ 2 (the angle of refraction ) we draw a triangle
Figure 3. Light comes in from the air on the left with the left dashed blue line indicating one wavecrest with the previous wavecrest having just finished passing into the water. Thus the wavelength λ 1 in medium 1 is the solid thick orange line on the bottom left, and the wavelength in the second medium λ 2 is the thick solid orange line on the top right.
Call the distance between the places where the wave crest hits the water along the water R (the thick green vertical line in the picture). The distance between crests is λ 1 = R sinθ 1 in the air and λ 2 = R sinθ 2 in the water. Since R is the same, trigonometry and n 1 λ 1 = n 2 λ 2 imply
n 1 n 2
λ 2 λ 1
sinθ 2 sinθ 1
This is known as Snell’s law. The same logic holds for reflected waves: R is the same and λ is the same (since n 1 = n 2 for a reflection) therefore θ 1 =^ θ 2. This is usual the^ law of reflection : the angle of reflection is equal to the angle of incidence. For a fast-to-slow interface (like air to water), the angle gets smaller (the refracted angle is less than the incident angle). For a slow-to-fast interface (like water to air), the angle gets larger. Since the angle cannot be larger than 90◦^ while remaining in the second medium, there is a largest incident angle for refraction when n 2 < n 1. In equations, since the refracted angle satis- fies sinθ 2 = n n^1 2 sinθ 1 , we see that if n n^1 2 sinθ 1 > 1 there is no solution. The critical angle beyond
which no refraction occurs is therefore
θc = sin−^1 n 2 n 1 (2)
For air n ≈ 1 and for water n = 1.33 so θc = 49 ◦. For incident angles larger than the critical angle, there is no refraction: all the light is reflected. We call this situation total internal reflection. Total internal reflection can only happen if n 2 < n 1. Thus, light can be confined to a material with higher index of refraction but not a lower one. Total internal reflection is the principle behind fiber optics. A fiber optical cable has a solid silica core surrounding by a cladding with an index of refraction about 1% smaller. For example, the core might have n 1 ≈ 1.4475 and the cladding n 2 = 1.444, so the critical angle is θc = 86 ◦^ from normal incidence, or 4 ◦^ from the direction of propagation. As long as the cable is not bent too much (typically the cable is thick enough so that this is very difficult), the light will just bounce around in the cable, never exiting, with little loss. Why might you want to send signals with vis- ible light rather than radio waves?
2 Section 1
For the magnetic field, the boundary conditions can be affected due to the magnetic moment of the particles in the medium, as encoded in μ. This picture looks like
Figure 5. Current on the boundary can affect B‖ not B⊥.
Since a current induces a field perpendicular to the current, only B‖ can be affected, not B⊥. Thus,
B⊥^ (1)^ = B⊥^ (2)^ (5)
Since the part of the magnetic field which is sensitive to an accumulated current in a material is
treated using H~^ = (^1) μ B~^ the condition is then that
1 μ 1
B||^ (1)^ =
μ 2
B||^ (2)^ (6)
In summary, the boundary conditions are
ǫ 1 E⊥^ (1)^ = ǫ 2 E⊥^ (2)^ B⊥^ (1)^ = B⊥^ (2) E||^ (1)^ = E||^ (2)^
μ 1
B||^ (1)^ =
μ 2
B||^ (2)^
In the vacuum, μ = μ 0 = 1.25× 10 −^6 mH (Henries per meter). If μ is much larger than this, the material can acquire a large current with a small magnetic field. That is, it conducts. For example, iron has μ = 6.3× 10 −^3. Conductors are opaque and therefore not of much interest for the study of refraction. In most transparent materials, μ ≈ μ 0. For example, water has μ = 1.256 × 10 −^6 Hm which is the same as for air up to 6 decimal places. In transparent materials, the electric permittivity can vary significantly. So we will assume that μ ≈ μ 0 and ǫ varies. In this
case, the boxed equations reduce to ǫ 1 E⊥^ (1)^ = ǫ 2 E⊥^ (2), E||^ (1)^ = E||^ (2)^ and B~^ (1)^ = B~^ (2).
Next, we’ll solve the wave equation with these boundary conditions, much like we solved the transmission and reflection problem for waves in a string in Lecture 9 which led to the concept of impedance.
3 Normal incidence
Let’s start with the case of normal incidence (perpendicular to the interface). To find out how much light is reflected, we need to work out the impedance, which determines the reflection and transmission coefficients.
4 Section 3
For normal incidence the incident angle to the normal θ 1 = 0 (see Fig. 8 below). Thus E⊥ = B⊥ = 0. That is, the electric and magnetic fields are both polarized in the plane of the interface to there simply is no ⊥ component. Without loss of generality, let’s take a plane wave of fre-
quency ω moving in the zˆ direction with E~^ in the yˆ direction. Then since B~^ = 1 ω k
~ (^) × E~ (^) , the mag-
netic field points in the xˆ direction. So the incident fields are
E~^ I = EIyˆe
iω
( t− (^) v^11 z
) , B~^ I = BIxˆe
iω
( t− (^) v^11 z
) (8)
The transmitted and reflected waves are also moving normal to the surface (by Snell’s law), so we can write
E~^ T = ETyˆe
iω
( t− (^) v^11 z
) B~^ T = BTxˆe
iω
( t− (^) v^11 z
) (9)
E~^ R = ERyˆe iω
( t+ (^) v^11 z
) B~^ R = −BRxˆe iω
( t+ (^) v^11 z
) (10)
with ET , ER, BT and BR the transmission and reflection coefficients to be determined. Note that we have flipped the sign on the z term in the phase since the reflected wave is moving in
the −zˆ direction (that is, ~k^ flips). Since ωB~^ = k~^ × E~^ , this means that B~^ flips sign too, which explains the minus sign in Eq. ( 10 ). Since there is no ⊥ component the boundary condition E||^ (1)^ = E||^ (2)^ implies that
EI + ER = ET (11)
Similarly, 1 μ 1 B||
μ 2 B||
(2) (^) implies
1 μ 1
BI −
μ 1
BR =
μ 2
BT (12)
Since
∣B~
v
∣E~
∣ (^) for any plane wave, BI = 1 v 1 EI^ and^ BT^ =^
1 v 2 ET^ and Eq. (^12 ) becomes 1 μ 1 v 1
(EI − ER) =
μ 2 v 2
ET (13)
Eqs. ( 11 ) and ( 13 ) look just very much like the equations for transmission and reflection for a wave that we studied in Lecture 9. Solving them gives
ER = EI
Z 2 − Z 1
Z 2 + Z 1
, ET = EI
2 Z 2
Z 2 + Z 1
where
Z = μv = μ c n
μ ε
For most materials, μ is pretty constant, so the form Z = μc (^) n^1 , which says Z ∝ (^1) n , is most useful
What is the power reflected and transmitted? The incident power in a medium is given by
the Poynting vector P~^ = (^1) μ E~^ × B~^ times area A. So
∣ ∣P~I
μv
∣E~ I
∣^2 A =^1
Z
∣E~ I
∣^2 A (16)
Thus the reflected power is
PR =
ER^2
Z 1
A =
Z 2 − Z 1
Z 2 + Z 1
EI^2
Z 1
A =
Z 1 − Z 2
Z 1 + Z 2
PI (17)
and the transmitted power is
PT =
ET^2
Z 2
A =
2 Z 2
Z 2 + Z 1
EI
Z 2
2 A =
4 Z 1 Z 2
(Z 1 + Z 2 )^2
EI^2
Z 1
A =
4 Z 1 Z 2
(Z 1 + Z 2 )^2
PI (18)
These satisfy PT + PR = PI as expected. Note that these equations hold for normal incidence only.
Normal incidence 5
In the limit that d ≪ λ the two reflections will be exactly out of phase. For T ≈ 1 , there will be complete destructive interference and no reflection. But there will also be complete destruc- tive interference whenever cos
4 π d λ
= 1, which is when
d = λ 2
2 λ 2
3 λ 2
4 λ 2 ··· ( complete destructive interference) (23)
On the other hand if the two waves are completely in phsae there will be constructive interfer-
ence. This happens when cos
4 π d λ
= − 1 which is when
d = λ 4
3 λ 4
5 λ 4 , ··· ( complete constructive interference) (24)
If the material is much thicker than the wavelength of light, and not of completely uniform thickness, then there will be some constructive and some destructive interference and we won’t see much interesting. However, if the material has a well-defined thickness which is of the same order of magnitude as the wavelength of visible light, we will see different wavelengths with dif- ferent intensities. This happens in a soap film.
If we put a soap film vertically, then gravity will make it denser at the bottom. The result is the following:
Note at the very top, the film is black because there is complete distructive interference for all wavelengths. Similar patterns can be seen in soap bubbles.
Color due to thin-film interference is known as iridescence. The color of many butterflies and the gorgets of hummingbirds is due to iridescence.
Thin film interference 7
Figure 7. Hummingbirds an butterflies get some of their color from thin-film interference.
5 Fresnel coefficients
Now let’s consider the more general case. Suppose we have a plane wave moving in the k~^ direc- tion towards a surface with normal vector ~n. Thus the angle θ 1 that the wave is coming in at satisfies n~ · k~^ =
∣~k
∣|~n |cos θ 1. The two vectors ~k (^) and ~n form a plane. There are two linearly inde-
pendent possibilities for the polarization: the electric field can be polarized in the ~k^ - ~n plane of the boundary, or transverse to that plane. We call these vertical and horizontal polariza- tions respectively. Snell’s law then tells us the directions of the transmitted and reflected elec- tric fields. The magnetic field is always determined by the electric field through ωB~^ = ~k^ × E~^. What we need to solve for is the amplitude of the transmitted and reflected fields. The two cases look like:
Vertical polarization Horizontal polarization (E in plane) (B in plane) Figure 8. Two linearly independent linear polarizations are vertical and horizontal. Crosses indicate vectors point into the page, and dots that the vectors come out.
8 Section 5
6 Transmitted power
To interpret the Fresnel coefficients, we would like to know not just the amplitude of the wave transmitted, but the intensity of the light that passes through, or equivalently the power. Recall
that the power in a plane wave in the vacuum is P = cǫ 0
∣E~
∣^2. For a plane wave in a medium, ε
changes and only the component of the velocity moving into the medium is relevant, so this
becomes P = v cosθǫ
E~^
= cosθ εμ
∣E~ ∣∣^2 = cosθ Z
∣E~ ∣∣^2. Thus the fraction of power reflected for ver-
tical and horizontal polarizations is:
PR^ vert PI^ vert^
α − β α + β
PR^ horiz PI^ horiz^
αβ − 1 αβ + 1
For the fraction of power transmited,
PT^ vert PI^ vert^
cosθ 2 cosθ 1
Z 1
Z 2
α + β
= αβ
α + β
and PT^ horiz PI^ horiz^
cosθ 2 cosθ 1
Z 1
Z 2
αβ + 1
= αβ
αβ + 1
One can check that for either polarization, PT + PR = PI and that these equations agree with
Eqs. ( 17 ) and ( 18 ) for normal incidence
α = 1, β = Z Z^1 2
For example, the air-glass interface has β = 1.5. Then the transmitted and reflected power in vertical and horizontal polarizations are
Figure 9. Transmitted and reflected power as a function of incident angle for the two polarizations.
We can also plot the ratio of power reflected in vertical to horizontal polarizations:
Figure 10. Ratio PR^ vert^ /PR^ h o r iz^ of power reflected vertically polarized to horizontally polarized light,.
From these plots we see there is an angle where the vertical polarization exactly vanishes. This angle is called Brewster’s angle , θB. We can see from the plots that for the glass-air interface θB ≈ 56 ◦. At this angle, the reflected light is completely polarized.
10 Section 6
What is the general formula for θB? From Eq. ( 37 ) we see that PR^ vert^ = 0 when α = β. That is cosθ 2 cosθB
ǫ 2 μ 1 ǫ 1 μ 2
For most materials μ 1 ∼ μ 2 ≈ μ 0. Then we can use n = μǫ
to get (^) coscosθθ^2 B = n n^21. Using also n 1 sinθB = n 2 sinθ 2 we can solve for θB giving
tanθB = n 2 n 1 (41)
For the air-water interface, θB = tan−^1 1.5 = 56.3◦. One can see from Figure 10 that one does not have to be exactly at this angle to have little reflected vertical polarization. Angles close to θB work almost as well. What is going on physically at Brewster’s angle? We know that Snell’s law n 1 sinθ 1 = n 2 sinθ 2 is always satisfied. At Brewster’s angle, when θ 1 = θB, we found also n 1 cosθ 2 = n 2 cosθ 1. Dividing these two equations gives cosθ 1 sinθ 1 = cosθ 2 sinθ 2 , or equivalently
sin(2θ 1 ) = sin(2θ 2 ) when θ 1 = θB (42)
Now, by definition θ 1 and θ 2 are both between 0 and π 2 , so 2 θ 1 and 2 θ 2 are between 0 and π. Thus, sin(2θ 1 ) = sin(2θ 2 ) has two solutions. Either θ 1 = θ 2 , which corresponds to n 1 = n 2 so the light just passes through, or π 2 − 2 θ 1 = 2θ 2 − π 2 (see Fig 11 , left) which simplifies to
θ 1 + θ 2 = π 2
Figure 11. The Brewster’s angle happens when sin(2θ 1 ) = sin(2θ 2 ) which corresponds to π 2 − 2 θ 1 = 2θ 2 − π 2 as can be seen in the left figure. This means that the transmitted and reflected waves are perpendic- ular. Thus the reflected vertically polarized light (shown on the right) cannot be produced by the motion of particles in the surface. Hence the reflected vertically polarized light vanishes at Brewster’s angle.
Working out the geometry, as in Fig. 11 , we see that the transmitted and reflected waves are perpendicular at Brewster’s angle. Since the reflected wave has to be produced by the motion of particles in the surface we can understand why there is no reflection at Brewster’s angle: parti- cles moving in the surface can only produce light polarized transverse to their direction of motion. In Lecture 17, we will understand in more detail how accelerating changes produce elec- tromagnetic fields and waves. The famous Harvard dropout Edwin Land started a corporation called Polaroid that made its first fortune with polarizing sunglasses. These glasses remove glare from reflection by removing the horizontally polarized light. He made his second fortune was made with the Polaroid camera. In 1973 Land donated the money for the construction of the Science Center (which some people say looks like a camera...).
Transmitted power 11
