Aoptics-1-wave fronts

Dr. G. Mirjalili, physics Dept. Yazd University

Applied optics

Wave fronts


Geometrical OpticsIn describing the propagation of

light as a wave we need to understand:

wavefronts: a surface passing through points of a wave that have the same phase and amplitude.

rays: a ray describes the direction of wave propagation. A ray is a vector perpendicular to the wavefront.


Wavefronts• We can chose to associate

the wavefronts with the instantaneous surfaces where the wave is at its maximum.

• Wavefronts travel outward from the source at the speed of light: c.

• Wavefronts propagate perpendicular to the local wavefront surface.


Light Rays

• The propagation of the wavefronts can be described by light rays.

• In free space, the light rays travel in straight lines, perpendicular to the wavefronts.


The ray approximation in geometric optics

• Geometric optics: The study of the propagation of light.

• Ray approximation: In the ray approximation, we assume that a wave moving through a medium travels in a straight line in the direction of its rays.


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


• 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


Reflection

• Reflection: When a light ray traveling in one medium encounters a boundary with another medium, part of the incident light is reflected.– Specular reflection: Reflection of light from a

smooth surface, where the reflected rays are all parallel to each other.

– Diffuse reflection: Reflection from any rough surface, where the reflected rays travel in random directions.

– we use the term reflection to mean specular reflection.


Reflection and refraction

Reflection (cont’d)


The Law of reflection

• Law of reflection: The angle of reflection equals the angle of incidence: 1

’ = 1.• Some definitions:

– Normal: The normal is a line drawn perpendicular to the surface at the point where the incident ray strikes.

– Angle of reflection and incidence: Measured from the normal to the reflected and incident rays, respectively.


Example : The double-reflected light ray

• Two mirrors make an angle of 120° with each other. A ray is incident on mirror M1 at an angle of 65° to the normal. Find the direction of the ray after it is reflected from mirror M2.

=2


Practical applications of reflection

• Retroreflection: If the angle between the two mirrors is 90°, the reflected beam will return to the source parallel to its original path.


Refraction• Refraction: When a ray of light traveling through

a transparent medium encounters a boundary leading into another transparent medium, part of the ray enters the second medium. The part that enters the second medium is bent at the boundary and is said to be refracted.

• sin2 / sin1 = v2 / v1

1 and 2 are the angle of incidence and angle of refraction, respectively.

– v1 and v2 are the speed of the light in the first and second medium, respectively.

• The path of a light ray through a refracting surface is reversible. All rays and the

normal lie in the same plane.


Reflection by plane surfacesr1 = (x,y,z)

x

y

r2 = (x,-y,z)

Law of Reflection

r1 = (x,y,z) → r2 = (x,-y,z)

Reflecting through (x,z) plane

x

y

zr2= (-x,y,z)

r3=(-x,-y,z)

r4=(-x-y,-z)

r1 = (x,y,z)


n2

Refraction by plane interface& Total internal reflection

n1

n1 > n2

θC

P

θ1θ1

θ1 θ1

θ2

θ2

Snell’s law n1sinθ1=n2sinθ2


Examples of prisms and total internal reflection

45o

45o

45o

45o

Totally reflecting prism

Porro Prism


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



Optical fibers


Index of Refraction and Snell’s Law of Refraction

• Index of refraction n of a medium: n c/v– c = 3 x 108 m/s: speed of light in vacuum.– v: speed of light in the medium; v < c.– n > 1 for any medium and n = 1 for vacuum (or

approximately in air).• Snell’s law of refraction: n1sin1=n2sin2 • As light travels from one medium to another, its

frequency does not change but its wavelength does. 1n1 = 2n2, or 1/2 = v1/v2.

• Light slows on entering a medium – Huygens

• Also, if n → ∞ = 0

• i.e. light stops in its track !!!!!


n1 = 1.5 n2 = 1.0

Reflections, Refractive offset

• Let’s consider a thick piece of glass (n = 1.5), and the light paths associated with it

– reflection fraction = [(n1 – n2)/(n1 + n2)]2

– using n1 = 1.5, n2 = 1.0 (air), R = (0.5/2.5)2 = 0.04 = 4%

incoming ray(100%)

96%

92% transmitted0.16%

4%

4%

8% reflected in tworeflections (front & back)

image looks displaceddue to jog


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


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.)

s`is reduced distance

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


• 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


Prisms

Applications of prism

• A prism and the total reflection can alter the direction of travel of a light beam.“Diversion, Deviation”

• All hot low-pressure gases emit their own characteristic spectra. A prism spectrometer is used to identify gases.“Dispersion”


Dispersion &Deviation

Little dispersion

High deviation

High dispersion

Low

deviation

n1 n2

n1<n2


Angular Dispersion

A hollow 600 prism is filled with carbon disulfide, whose index of refraction for blue is 1.652, for red light is 1.618 what is angular dispersion

2

1sin

)(2

1sin

0

n

n prism

n1=1.652 n2=1.618

1=51.380 2=48.00

1-2 =3.380 angular dispersion


Deviation & wavenumber in prism

• Deviation angle &

b+nt1+d=c+nt2+e

b+(n+ n)t1+d+a D=c+ (n+ n)t2+e

nt1+a D=n t2

D/ =(t2-t1)/a

A

B

A`b

c

d

e

t1

t2

A``

aD

a

dD/dn = (t2-t1)/a

dD/dn =t/a

dD/d=(dD/dn)(dn/d)=(t/a) (dn/d)

(dn/d)=?


Deviation angle &

n = A+B/2

dn/d = -2B/3

dD/d = t/a dn/ddD/d = t/a(-2B/3)


Dispersive power & Abbe`s nunber

1

D

cf

n

nnDispersive power

numbersAbbenn

n

Cf

D `1

Low dispersion, low refractive index


Refractive indices of Crown and flint glasses

Fraunhofer line

color (nm) n crown n flint

F

D

C

Blue

Yellow

Red

486.1

589.3

656.3

1.5293

1.5230

1.55204

1.7378

1.7200

1.7130

Crown =59 flint=29


Dispersing prisms

• Achromatic prism:

• Deviates light but gives no dispersion

1

2

Dispersion for 1 and 2 is zero


Dispersing prisms

• Direct-vision prism

1

2

Direct vision for wavelength


example• Assume that 140 is the apex angle of a crown glass prism. What

should be the apex angle of a flint prism:(a)-if the combination of both is to be achromatic for blue and red?(b)-if the prism is to have no deviation for yellow?Solution (a)

F = 1(nf - 1)

C = 1(nc- 1) → f - c = 1 (nf - nc) mean dispersion of prism

For the combination to be achromatic,

(1)(n1f - n1c) + (2)(n2f - n2c)=0

(14)(1.5293-1.5204)+ (2)(1.7378-1.7130)=0

2= -50


Example (cont.)

• (b) for the direct-vision prism

1=1(n1D-1)

2=2(n2D-1)

1(n1D-1)= 2(n2D-1)

14(1.5230-1)= 2(1.7200-1)

2=10.20


Image manipulation by reflection prisms

Right angle prism


Image manipulation by reflection prisms

Dove prism


Dispersion

Dispersion

• The index of refraction of a material depends on wavelength as shown on the right. This is called dispersion.

• 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 & dispersion

Examples


Resolving power of a prism

+

d

b

Ts

FT+TW=nb

FT+ Tw - s= (n- n) b

s=b n

s=b (dn/d)

=s/d=(b/d)(dn/d)

F

d W

n

= /d

/d = (b/d)(dn/d)

()min= /b(dn/d)

R=/()min= b(dn/d)


Resolving power of a prism (example)

• A prism made from flint glass with a base of 5 cm. find the resolving power of the prism at =550 nm.

• solution

n/=(nf-nD)/(f-D)= (1.7328-1.7205)/(486-587)=-1.9x10 -4 nm -1

R = b(dn/d) = (0.05x10 9nm)(-1.9x10 -4 nm -1) = 5971

()min=/R =5500A0/5971 ≈ 1 A0


•Exercises


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

Aoptics-1-wave fronts

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