Date post: | 04-Jan-2016 |
Category: |
Documents |
Upload: | olivia-caldwell |
View: | 222 times |
Download: | 1 times |
Speed of light (in vacuum)
Foucault’s experiment
Michelson’s 1878 Rotating Mirror Experiment• 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.
Speed of light (in vacuum)
Nature of light
Waves, wave fronts, and rays
• Wave front: The locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same.
rays
wave fronts
source
spherical wave plane wave
Reflection and refraction
Reflection and refraction
• When a light wave strikes a smooth interface of two transparent media (such as air, glass, water etc.), the wave is in general partly reflected and partly refracted (transmitted).
incident raysreflected rays
refracted rays
ar
b
bb
a a
Reflection and refraction
Reflection
incident raysreflected rays
refracted rays
ar
b
b
a
• The angle of reflection is equal to the angle of incidence for all wavelengths and for any pair of material.
r
r a
a
• The incident, reflected, and refracted rays, and the normal to the surface all lie in the same plane.
Reflection and refraction
Reflection (cont’d)
Reflection and refraction
Refraction
• The index of refraction of an optical material (refractive index), denoted by n, is the ratio of the speed of light c in vacuum to the speed v in the material.
nvcn /;/ 0 wavelength in vacuum. Freq. stays the same.
f=v
Reflection and refraction
Refraction
incident raysreflected rays
refracted rays
ar
b
b
a
• The ratio of the sines of the angles and , where both angles are measured from the normal to the surface, is equal to the inverse ratio of the two indices of refraction:
a b
• The index of refraction of an optical material (refractive index), denoted by n, is the ratio of the speed of light c in vacuum to the speed v in the material.
a
b
b
a
n
n
sin
sinSnell’s law
nvcn /;/ 0 wavelength in vacuum. Freq. stays the same.
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.)
L
L
21
2
1
21
tan)(tan
'tan
tan
sinsin
sss
s
L
ss
Ls
L
na
for small angles: tan ->sin
.504
1)2(
1
sin)(sin
sin)(sin
11
21
cmmn
nss
nsss
sss
a
a
a
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
sn
ns
s
n
s
n
1
221 ''
L
)sinsin( '
sinsin'
1for sintan
tantan'tan||,tan|'|
221121
21
2121
nnsnsn
ss
ssLsLs
s’
s
Total internal reflection
Total internal reflection
,sinsin 21
21 n
nSince 1sin 2 when .sin&1/ 11212 nnnn
When this happens, is 90o and is called critical angle. Furthermore2 1when , all the light is reflected (total internal reflection). crit 1
Total internal reflection
Optical fibers
• Light is refracted twice – once entering and once leaving. • Since n decreases for increasing , a spectrum emerges...
Analysis: (60 glass prism in air)
1 2
4
n2 = 1.5
n1 = 1 60
sin 1 = n2 sin 2
n2 sin 3 = sin 4
3
3 = 90 - = 90 - 2 3 = 60 - 2
Example: 1 = 30
o
oo
o
9.76sin5.1sin
5.40)60(
5.195.1
)30sin(sin
31
4
23
12
o = o
Prism example
Prism
Applications of prism
• A prism and the total reflection can alter the direction of travel of a light beam.
• All hot low-pressure gases emit their own characteristic spectra. A prism spectrometer is used to identify gases.
Diversion
Diversion
• The index of refraction of a material depends on wavelength as shown on the right. This is called diversion.
• It is also true that, although the speed of light in vacuum does not depends on wavelength, in a material, wave speed depends on wavelength.
Diversion
Examples
Huygens’ principle Huygens’ principle
Every 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
• 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 st, a new plane BB’ can be drawn tangent to the wavefronts
Huygens’ principle (cont’d) Huygens’ principle for plane wave
Huygens’ principle (cont’d) Huygens’ principle for spherical wave
• 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 spherical wave (cont’d)
• The law of reflection can be derived from Huygen’s Principle
• AA’ is a wave front of incident light
• The reflected wave front is CD
Huygens’ principle (cont’d) Huygens’ principle for law of reflection
• Triangle ADC is congruent to
triangle AA’C
• Angles 1 = 1’
• This is the law of reflection
• 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 1 = n2 sin 2
Huygens’ principle (cont’d) Huygens’ principle for law of refraction
22
11
2
2
1
1
2211
,,sinsin
sin;sin
n
cv
n
cv
tvtv
tvtv
AC
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
ExercisesExample
Solution
m
n
A
A
The 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?
.6.44)00.1
0.25sin66.1(sin)
sin(sinsinsin 11
b
aabbbaa n
nnn
Therefore the angle below the horizon isand thus the angle between the two emerging beams is
,6.190.256.440.25 b.2.39
ExercisesExample
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
a
'a
b
'b
.'aa
,cos
)sin('
'
b
batd
ExercisesProblem
Solution
P
Q
n
n’
n
t
d
a
'a
b
'b(a)For light in air incident on a parallel-faced
plate, Snell’s law yields:
(b) Adding more plates just adds extra steps in the middle of the above equation that always cancel out. The requirement of parallel faces ensures that the angle and the chain of equations can continue.(c) The lateral displacement of the beam can be calculated using geometry:
(d)
.sinsinsinsin'sin'sin ''''aaaaabba nnnn
'nn
.cos
)sin(
cos),sin(
'
'
''
b
ba
bba
td
tLLd
L
.62.15.30cos
)5.300.66sin()40.2(
5.30)80.1
0.66sin(sin)
'
sin(sin 11'
cmcm
d
n
n ab