ri θθ =

2211 sinsin θθ nn =

Index of Refraction

• Speed of light, c, in vacuum is 3x108m/s

• Speed of light, v, in different medium can be v < c.

• index of refraction, n = c/v.

• frequency, f, does not change in wave eqn. of v = f λ,

• wavelength, λ, depends on medium, λ = v/f = c/nf = λ0/n

• In some media, n, depends on f, this is called dispersion.

Waves, wave front, rays (Y&F section 33.2)Plane waves moving in +x direction

( ) ( )( ) ( )tkxzBtzyxB

tkxyEtzyxEω

ω

−=

−=

cosˆ,,,cosˆ,,,

0

0r

r

x

y

E field Wave front (y-z plane surface of constant phase)

Ray propagation

Wave front is a surface of constant phase

z

Huygen’s Principle (Y&F section 33.7)• Huygen’s principle; a wave front can be a source of secondary wavelets the spread out in all directions at the speed of propagation in the medium. The envelope of leading edges forms a wave front.• This principle was stated by Huygen in 1678, it is derivablefrom Maxwell’s eqn. It is a geometrical description of raypropagation.

Plane wave example;

Secondary wavelets create another wave front (plane)

vt

Reflection from Huygen’s Principle

θi

vt

vtvt vt

θr

Incident wave frontreflected wave front

Consider wave fronts, separated by vt, the incident wave fronts in contact with the surface will create a wavelets according to Huygen’sPrinciple and leds to another “reflected” wave front. Result is θi = θr

Reflection Lawθi = θr

θi θr

ray diagram

Snell’s Law (law of refraction)

θaθb

= nb sinθbna sinθa

Snell’s Law

θb

θaWave fro

nt

Refracted Wave front

Incident ray

Refracted rayMedium b, vb=c/nb

norm

al

Angles are usually givenrelative to normal to plane

Medium a, va=c/na

Refraction from Huygen’s Principle

θaθb vbt

L L

L sinθb = vbt = ct/nb

vatL sinθa = vat = ct/na

vat

vbt

Now the speed changes, from medium a to medium b, so theSpeed may change and the wavefront spacing differs.

= nb sinθbna sinθa

Snell’s Law

Medium a, va=c/na

Medium b,vb=c/nb

Wave front

Wave front

2) A ray of light passes from air into water with an angle of incidence of 30o. Which of the following quantities does not change as the light enters the water. Mark all correct answers.

a) Wavelength

b) frequency

c) speed of propagation

d) direction of propagation.

Preflight 24

EXAMPLE, from air into glass;

Suppose we have light in air (n=1)incidence on glass (n=1.55) at anangle θa=45 deg. What is the angleof the refracted light, θb?

θb

θa

( )( )

°=

=°

=

=°

=

27255.1

155.1

45sin)1(sin

sin)55.1(45sin)1(

sinsin

b

b

b

bglassaair nn

θ

θ

θ

θθ

EXAMPLE, from glass into air;Suppose we have light in glass (n=1.55) incident into air at an angle θa=30 deg. What is the angle of the refracted light, θb?

θb

θa

( )

( )

°=

=°=

=°

=

51255.145sin)55.1(sin

sin)1(30sin)55.1(

sinsin

b

b

b

bairaglass nn

θ

θ

θ

θθ

Lecture 24, ACT 1• Which of the following ray diagrams could represent the passage of

light from air through glass and back to air? (nair=1 and nglass=1.5)

(a) (b) (c)air

air

glass

air

air

glass

air

air

glass

Lecture 24, ACT 1• Which of the following ray diagrams could represent the passage of

light from air through glass and back to air? (nair=1 and nglass=1.5)

(a) (b) (c)air

air

glass

air

air

glass

air

air

glass

• Since n(glass) > n(air), sinθ (glass) < sinθ (air) .• Therefore, moving from air to glass, ray will bend toward normal.

• this eliminates (a).• Moving from glass to air, ray will bend away from normal.

• this eliminates (c).• As a matter of fact, the final angle in air must be equal to the initial angle in air!!

θ2

θ1

• The behavior of these rays is determined from Snell’s Law:

2211 sinsin θθ nn =

Suppose you are stranded on an island with no food. You see a fish in the water. Where should you aim your spearto hit the fish?

ANSWER; do not aim directly at the apparent position ofthe fish. Aim at the inside of the fish.

Suppose in the previous question instead of a spear youhad a high power laser to simultaneously kill and cook thefish (in the water). Where should you aim the laser??

ANSWER; aim directly at apparent fish position as the laser beam will refract to the correct fish position.

Total Internal Reflection– Consider light moving from glass (n1=1.5) to air (n2=1.0)

I.e., light is bent away from the normal.as θ1 gets bigger, θ2 gets bigger, but θ2can never get bigger than 90° !!

In general, if sin θ1 > (n2 / n1), we have NO refracted ray; we have TOTAL INTERNAL REFLECTION.

For example, light in water which is incident on an air surface with angle θ1 > θc = sin-1(1.0/1.33) = 48.8° will be totally reflected. This property is the basis for the optical fiber communication.

incident ray

reflected ray

refracted ray

θ2

θ1 θrGLASS

AIRn2

n1

2

1sinsin

2

1

1

2 >=nn

θθ

12 θθ >

4) The path of light is bent as it passes from medium 1 to medium 2. Compare the indexes of refraction in the two mediums.

a) n1 > n2

b) n1 = n2

c) n1 < n2

5) A light ray travels in a medium with n1 and completely reflects from the surface of a medium with n2. The critical angle depends on:

a) n1 only

b) n2 only

c) n1 and n2

Snell’s Law: n1sinθ1 = n2sinθ2Here, θ2 >θ1 implies n2 < n1

Critical angle occurs when θ2 = 90o

Therefore, sinθcritical = n2/n1

Preflight 24

Examples of devices using Critical Angle

• Prism Binoculars

• Fiber Optics

Fiber optics is extremely important for high speedInternet and digital data transfer at long distances.Many companies (Lucent) have laid fiber over long Distances to provide internet service.

Total Internal Reflection

Total internal reflection occurs when θ>θc and provides 100% reflection. This has better efficiency than silvered mirror.

I2 (academic) network of fiber connections

10 gigabit/secconnections

air n =1.00

glass n =1.5

air n =1.00

θc

water n =1.33

glass n =1.5

water n =1.33

θc

Case I

Case II

An optical fiber is surrounded by another dielectric. In case I this is water, with an index of refraction of 1.33, while in case II this is air with an index of refraction of 1.00.

Compare the critical anglesfor total internal reflection in these two cases

a) θcI>θcII

b) θcI=θcII

c) θcI<θcII

Lecture 24, ACT 2: Critical Angle…

air n =1.00

glass n =1.5

air n =1.00

θc

water n =1.33

glass n =1.5

water n =1.33

θc

Case I

Case II

An optical fiber is surrounded by another dielectric. In case I this is water, with an index of refraction of 1.33, while in case II this is air with an index of refraction of 1.00.

Compare the critical anglesfor total internal reflection in these two cases

a) θcI>θcII

b) θcI=θcII

c) θcI<θcII

Since n1>n2 TIR will occur for θ > critical angle. Snell’s law says sinθc=n2/n1.

If n2=1.0, then θc is as small as it can be.

So θcI >θcII .

n1>n2

n1

n2

Lecture 24, ACT 2: Critical Angle…

Dispersion: n = n(ω)In

dex

of re

fract

ion

frequency

ultravioletabsorption

bands1.50

1.52

1.54 white light

prism

Split into Colors

The index of refraction depends on frequency, due to the presence of resonant transition lines.

For example, ultraviolet absorption bands in glass cause a rising index of refraction in the visible, i.e., n(higher ω) > n(lower ω):

nred = 1.52 nblue = 1.53

Rainbows

Rainbows