Chapter 30: Reflection and Refraction The nature of light Speed of light (in vacuum) c = 2.99792458 x 10 8 m/s measured but it is now the definition
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Chapter 30: Reflection and Refraction
The nature of lightSpeed of light (in vacuum)
c = 2.99792458 x 108 m/s measured but it is now the definition
Michelson’s 1878 Rotating Mirror Experiment
Picture credit
• German American physicist A.A. Michelson realized, on putting together Foucault’s apparatus, that he could redesign it for much greater accuracy.
• Instead of Foucault's 60 feet to the far mirror, Michelson used 2,000 feet.. • Using this method, Michelson was able to calculate c = 299,792 km/s• 20 times more accurate than Foucault• Accepted as the most accurate measurement of c for the next 40 years.
The nature of light
Waves, wavefronts, and rays• Wavefront: The locus of all adjacent points at which the phase ofvibration of a physical quantity associated with the wave is the same.
rays
wavefronts
source
spherical wave plane wave
Reflection and refraction
Reflection and refraction• When a light wave strikes a smooth interface of two transparentmedia (such as air, glass, water etc.), the wave is in general partlyreflected and partly refracted (transmitted).
incident raysreflected rays
aθrθ
bθ
bb
a a
refracted rays
Reflection and refraction
Reflection
• The incident, reflected, and refracted rays, and the normal to thesurface all lie in the same plane.
incident raysreflected rays• The angle of reflection is equal tothe angle of incidence for allwavelengths and for any pair ofmaterial.
rθaθ
aθrθ
bθ
b
a
ar θθ =
refracted rays
Reflection and refraction
Refraction
incident raysreflected rays
• The index of refraction of an optical material (refractive index), denotedby n, is the ratio of the speed of light c in vacuum to the speed v in thematerial.
nvcn /;/ 0λλ ==wavelength in vacuum. Freq. stays the same.
aθrθ
bθ
b
• The ratio of the sines of the anglesand , where both angles are
measured from the normal to thesurface, is equal to the inverse ratioof the two indices of refraction:
aθ bθ
a
a
b
b
a
nn
=θθ
sinsin
Snell’s lawrefracted rays
Total internal reflection
Total internal reflection
,sinsin 21
21 θθ
nn
= 1sin 2 =θ whenSince .sin&1/ 11212 θnnnn =>
When this happens, is 90o and is called critical angle. Furthermore2θ 1θwhen , all the light is reflected (total internal reflection). critθθ >1
Total internal reflection
Optical fibers
Dispersion
Dispersion
• The index of refraction of amaterial depends on wavelengthas shown on the right. This iscalled dispersion. It is also true that,although the speed of light in vacuumdoes not depends on wavelength,in a material wave speed dependson wavelength.
Diversion
Examples
Huygens’ principleHuygens’ principleEvery point of a wave front may be considered the source of secondarywavelets that spread out in all directions with a speed equal to the speedof propagation of the wave.
Plane waves
Huygens’ principle (cont’d)Huygens’ principle for plane wave
• At t = 0, the wave front is indicated by the plane AA’
• The points are representative sources for the wavelets
• After the wavelets have moved a distance c∆t, a new plane BB’ can be drawn tangent to the wavefronts
Huygens’ principle (cont’d)Huygens’ principle for spherical wave
Huygens’ principle (cont’d)Huygens’ principle for spherical wave (cont’d)
• The inner arc represents part of the spherical wave
• The points are representative points where wavelets are propagated
• The new wavefront is tangent at each point to the wavelet
Huygens’ principle (cont’d)Huygens’ principle for law of reflection
• The Law of Reflection can be derived from Huygen’s Principle
• AA’ is a wavefront of incident light
• The reflected wave front is CD
Huygens’ principle (cont’d)Huygens’ principle for law of reflection (cont’d)
• Triangle ADC is congruent to triangle AA’C• Angles θ1 = θ1’• This is the law of reflection
Huygens’ principle (cont’d)Huygens’ principle for law of refraction
• In time ∆t, ray 1 moves from A to B and ray 2 moves from A’ to C
• From triangles AA’C and ACB, all the ratios in the law of refraction can be found:
n1 sin θ1 = n2 sin θ2
AC=l
22
11
2
2
1
1
2211
,,sinsin
sin;sin
ncv
ncvtvtv
tvtv
==∆
=∆
→
∆=∆=
θθ
θθ ll
Polarization
EM wave
BE
tkzBjtzB
tkzEitzE
rr
r
r
⊥
−=
−=
)cos(ˆ),(
)cos(ˆ),(
max
max
ω
ω
Polarization (defined by the direction of )Er
Linear polarization
z
x
y
In the text:E(x,t)=jEmaxcos(kx-ωt)B(x,t)=kBmaxcos(kx-ωt)^
^
Circular polarization
Polarization (cont’d)Polarization (defined by the direction of )E
r
Circular polarization
Polarization (cont’d)Polarizing filters
Polarization (cont’d)Polarization by reflection
pθ pθ
bθ
an
bn
plane of incidenceWhen the angle of incident coincides withthe polarizing angle or Brewster’s angle,the reflected light is 100% polarized.
pθ
pbpbbbpa nnnn θθθθ cos)90sin(sinsin =−°==
a
bp n
n=θtan Brewsters’s law of the polarizing angle
Example: depth of a swimming pool
Pool depth s = 2m
person looks straight down.
the depth is judged by the apparent size of some object of length L at the bottom of the pool (tiles etc.)
θ2
θ1
L
21
2
1
21
tan)(tan
'tan
tan
sinsin
θθ
θ
θ
θθ
sss
sL
ssL
sL
na
∆−=→
=∆−
=
=
=
for small angles: tan ->sin
θ2
θ1
L .5041)2(1
sin)(sinsin)(sin
11
21
cmmn
nss
nssssss
a
a
a
==−
=∆
∆−=∆−=
θθθθ
Example: Flat refracting surface
• The image formed by a flat refracting surface is on the same side of the surface as the object– The image is virtual– The image forms between
the object and the surface– The rays bend away from
the normal since n1 > n2
Lθ2
θ1 θ2
pnnq
qn
pn
1
221 =⇒=
)sinsin(
sinsin1for sintan
tantantan||,tan||
221121
12
1212
θθ
θθθθθθ
θθθθ
nnpnqn
pq
pqLpLq
==⇒
=⇒
<<≈≈
=→==
Q
Prism example• Light is refracted twice – once entering and once leaving. • Since n decreases for increasing λ, a spectrum emerges...
Analysis: (60° glass prism in air)
1θ 2θ
4θ
n2 = 1.5
n1 = 1 60°
sin θ1 = n2 sin θ2
n2 sin θ3 = sin θ4
3θ
αβ
θ3 = 90° - β α = 90° - θ2 → θ3 = 60° - θ2
Example: θ1 = 30°
( ) o
oo
o
9.76sin5.1sin
5.40)60(
5.195.1
)30sin(sin
31
4
23
12
==
=−=
=⎟⎠⎞
⎜⎝⎛=
−
−
θθ
θθ
θ
α+β+60o = 180o
Atmospheric Refraction and Sunsets
• Light rays from the sun are bent as they pass into the atmosphere
• It is a gradual bend because the light passes through layers of the atmosphere – Each layer has a slightly
different index of refraction• The Sun is seen to be above
the horizon even after it has fallen below it
Mirages
• A mirage can be observed when the air above the ground is warmer than the air at higher elevations
• The rays in path B are directed toward the ground and then bent by refraction
• The observer sees both an upright and an inverted image
ExercisesProblem 1The prism shown in the figure has a refractiveindex of 1.66, and the angles A are 25.00 . Twolight rays m and n are parallel as they enterthe prism. What is the angle between themthey emerge?
Solution
m
n
A
A
.6.44)00.1
0.25sin66.1(sin)sin(sinsinsin 11 °=°
==→= −−
b
aabbbaa n
nnn θθθθ
Therefore the angle below the horizon isand thus the angle between the two emerging beams is
,6.190.256.440.25 °=°−°=°−bθ.2.39 °
ExercisesProblem 2
Light is incident in air at an angle on the upper surface of a transparentplate, the surfaces of the plate beingplane and parallel to each other. (a)Prove that (b) Show that thisis true for any number of different parallelplates. (c) Prove that the lateral displacementd of the emergent beam is given by therelation:
where t is the thickness of the plate. (d) A ray of light is incident at an angleof 66.00 on one surface of a glass plate 2.40 cm thick with an index ofrefraction 1.80. The medium on either side of the plate is air. Find the lateralDisplacement between the incident and emergent rays.
P
Q
n
n’
n
t
d
aθ
'aθ
bθ
'bθ
.'aa θθ =
,cos
)sin('
'
b
batdθθθ −
=
ExercisesProblem 2
Solution
P
Q
n
n’
d
a
t
n
θ
'aθ
bθ
'bθ(a)For light in air incident on a parallel-faced
plate, Snell’s law yields:
(b) Adding more plates just adds extra stepsin the middle of the above equation thatalways cancel out. The requirement ofparallel faces ensures that the angleand the chain of equations can continue.
(c) The lateral displacement of the beam can be calculated using geometry:
(d)
.sinsinsinsin'sin'sin ''''aaaaabba nnnn θθθθθθθθ =→=→===
'nn θθ =
.cos
)sin(cos
),sin(b
ba
bba
tdtLLdθθθ
θθθ −
=→=−=
L
.62.15.30cos
)5.300.66sin()40.2(
5.30)80.1
0.66sin(sin)'
sin(sin 11
cmcmd
nn a
b
=°
°−°=→
°=°
== −− θθ
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