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