Monochromatic plane waves ( ) Plane waves have straight wave fronts

Monochromatic plane waves
Plane waves have straight wave fronts
– As opposed to spherical waves, etc.
– Suppose
E ( r ) = E 0 e ik i r
Î
− iωt
E ( r, t ) = Re{E(r )e
λ
}
= Re{E0 eik i r e − iωt }
= Re{E0 ei ( k i r −ωt ) }
– E0 still contains: amplitude, polarization, phase
– Direction of propagation given by wavevector:
k = (k x , k y , k z ) where |k|=2π /λ =ω /c
– Can also define E = ( Ex , E y , Ez )
– Plane wave propagating in z‐direction
E ( z , t ) = Re{E0 ei( kz −ωt ) } = 12 {E0 ei( kz −ωt ) + E*0 e − i( kz −ωt ) }
Ex
By
vz
Linear versus Circular polarization
Polarization: Summary
ŷ
E = Exeiδ1 xˆ + Eyeiδ2 yˆ
ŷ E
x̂
x̂
linear polarization
y‐direction
right circular polarization
left circular polarization
Phase difference Î
90 0 (π/2, λ/4)
Phase difference = 00
Ex
Phase difference Î
180 0 (π, λ/2)
Ex
ẑ
Ey
ẑ
Ey
ẑ
left elliptical polarization
ẑ
Ex
ẑ
Ey
ẑ
Poynting vector & Intensity of Light S = E × H
•Poynting vector describes flows of E‐M power
•Power flow is directed along this vector
•Usually parallel to k
•Intensity is equal to the magnitude of the time averaged Poyning vector: I=<S>
cε 0 2 cε 0
S = I ≡| E ( t ) × H ( t ) |=
E =
Ex 2 + E y 2 )
(
2
2
cε 0 ≈ 2.654 ×10−3 A / V
example E = 1V / m
I = ? W / m2
ω[eV ] =
1239.85
λ[nm]
= 1.05457266 × 10 Js
−34
Maxwell’s equations in a medium (source‐free)
The induced polarization, P, contains the effect of the medium: ∇⋅E = 0
∇⋅H = 0
∂H
∇ × E = −μ
∂t
∂E ∂P
∂E
∇ × H = ε0
+
=ε
∂t ∂t
∂t
B = μH
D = ε0E + P = ε E
P = ε0χ E
The polarization is proportional to the field:
P = ε0χ E
This has the effect of simply changing the dielectric constant (refractive index n):
ε = ε 0 (1 + χ ) = n 2
Wave equations in a medium
The induced polarization in Maxwell’s Equations yields another term in the wave equation:
∂2 E 1 ∂2 E
− 2 2 =0
2
v ∂t
∂z
∂ E
∂ E
− με 2 = 0
2
∂z
∂t
2
2
This is the Inhomogeneous Wave Equation.
The polarization is the driving term for a new solution to this equation.
∂2 E
∂2 E
− μ 0ε 0 2 = 0
2
∂z
∂t
∂2 E 1 ∂2 E
− 2 2 =0
2
∂z
c ∂t
Homogeneous (Vacuum) Wave Equation
E ( z , t ) = Re{E0 ei( kz −ωt ) }
= 12 {E0 e
i ( kz −ωt )
+E e
*
0
− i ( kz −ωt )
=| E0 | cos ( kz − ωt )
}
c
=n
v
Phase velocity
Reflection and Transmission @ dielectric interface Snell’s Law
kxRegion1 = kxRregion2
n1k0sinθ1 = n2k0sinθ2
n1sinθ1 = n2sinθ2
The fish problem …
si/so = n’/n
si
so
s
so
i
Displacement by a glass plate
d=
(
t sin θ − θ '
( )
cos θ '
)
Reflection and refraction at a flat dielectric interface
Reflection and Transmission (Fresnel’s equations)
Can be deduced from the application of boundary conditions of EM waves.
Normal Incidence
Reflection and Transmission of Energy @ dielectric interfaces
Prisms
Thin Lens Imaging
Paraxial approximation
sin (θ ) ≈ tan (θ ) ≈ θ
cos (θ ) ≈ 1
See Hecht Ch. 5 and review the following Equations. “Sign” convention :
(See Hecht Table 5.1, Fig. 5.12, Table 5.2) 1 1 1
= +
f s0 si
x0 xi = f 2
MT ≡
Thick lens
n=nl/nm
yi
s
=− i
y0
so
dxi
f2
=− 2
ML ≡
dx0
x0
Thin lens (d is negligible)
Object
Image
Bi‐convex
Bi‐concave
⎡1
1 1 ( nl − nm ) ⎡ 1
1 ⎤
1 ⎤ 1
+ =
⎢ − ⎥ = ( nlm − 1) ⎢ − ⎥ =
so si
nm
⎣ R1 R2 ⎦
⎣ R1 R2 ⎦ f
⎡1
nm nm
1 ⎤
+
= ( nl − nm ) ⎢ − ⎥
so si
⎣ R1 R2 ⎦
nl is the lens index, nm is the medium index
⎡1
1 1
1 ⎤ 1
+ = ( nlm − 1) ⎢ − ⎥ =
so si
⎣ R1 R2 ⎦ f
A ray propagates from left to right, passing through surface 1, then surface 2
If R1 < R2, and R1, R2 >0, [ ] > 0 (convex)
If R1 > R2, and R1, R2 >0, [ ] < 0 (concave)
Positive Negative
Imaging by a single thin lens
Ray tracing s > f
2f
f
f
2f
f
f
2f
f
f
2f
f
f
2f
Positive lens: real, inverted image
magnification MT= y’/y = ‐s’/s = ‐si/so
2f
Magnifying glass
Ray tracing s < f
Positive lens: virtual, erect image
Magnifying glass
25cm
For reading distance
f
25cm
M=
For infinite image
f
M = 1+
θ
M=
θo
Negative lens –
Summary : Real and Virtual Images
ray tracing & virtual image
Lens Combination
1.
Form intermediate image from L1 at P’. 2.
Use intermediate image at P’ as input to L2.
Ray tracing – two lens combination
Lens distance d < f, i.e. f1 or f2
1. Draw rays 2 & 3 without the presence of L2. Form intermediate image at P’. 2. Draw ray 4 through O2. This is unchanged after L2 is inserted. 3. Draw ray 3 in the presence of L2 so that it passes through Fi2. Image appears at P1 .
Spherical Mirror
Hecht Ch. 5.4.3
Mirror Formula
1 1 1
= +
f s0 si
Paraxial Approximation
f =R/2
h0
f
hi
Si
S0
See Hecht Table 5.4 for sign convention
Aperture Stop and Entrance & Exit Pupil
Aperture Stop (AS)
The aperture stop (AS) is defined to be the stop or lens ring, which physically limits the solid angle of rays passing through the system from an on‐axis object point. The aperture stop limits the brightness of an image.
The entrance pupil of a system is the image of the aperture stop as seen from an axial point on the object through those elements preceding the stop. (Hecht p. 171)
The exit pupil of a system is the image of the aperture stop as seen from an axial point on the image plane through the interposed lenses, if there is any. (Hecht p. 172)
The Chief Ray
For an off‐axis object, the chief ray (CR) is the ray that passes through the center of the aperture stop. Rays that pass through the edge of the aperture stop are marginal rays (MR).
Example II: Aperture Stop + Field Stop
Telescope
• Object is at infinity so image is at f
• Measure angular magnification
• Length of telescope light path is sum of focal lengths of
objective and eyepiece
fo
M =−
fe
CA0 s θ '
= = = M.
CAe s' θ
The exit pupil is the image of the aperture stop (AS).
Define CA0 = entrance pupil clear aperture
CAe= exit pupil clear aperture
From the diagram, it is clear that
Microscope
• Magnification is product of lateral
magnification of objective and angular
magnification of eyepiece
• Note: Image is viewed at infinity
s2 − x '
h'
M0 = = − =
h
s1
f0
Me =
25
fe
M total
− x ' 25
= M0 × Me =
⋅
f0 fe