Aoptics-1-wave Fronts

Applied optics Wave fronts

Dr. G. Mirjalili, physics Dept. Yazd University

Geometrical Optics In 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.

Dr. G. Mirjalili, physics Dept. Yazd University

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. Dr. G. Mirjalili, physics Dept. Yazd University

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.

Dr. G. Mirjalili, physics Dept. Yazd University

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.

Dr. G. Mirjalili, physics Dept. Yazd University

Huygens’ principle  Huygens’ principle Every point of a wave front may be considered the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the wave. Plane waves

Dr. G. Mirjalili, physics Dept. Yazd University

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 s∆ t, a new plane BB’ can be drawn tangent to the wavefronts

Dr. G. Mirjalili, physics Dept. Yazd University

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

Dr. G. Mirjalili, physics Dept. Yazd University

Huygens’ principle (cont’d)  Huygens’ principle for law of reflection •

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

• Triangle ADC is congruent to triangle AA’C • Angles θ 1 = θ 1’ • This is the law of reflection Dr. G. Mirjalili, physics Dept. Yazd University

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 θ 1 = n2 sin θ 2

sin θ1 = v1∆t; sin θ 2 = v2 ∆t →

v1∆t v ∆t c c = 2 , v1 = , v2 = sin θ1 sin θ 2 n1 n2 Dr. G. Mirjalili, physics Dept. Yazd University

 = 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. Dr. G. Mirjalili, physics Dept. Yazd University

Reflection and refraction  Reflection (cont’d)

Dr. G. Mirjalili, physics Dept. Yazd University

The Law of reflection • Law of reflection: The angle of reflection equals the angle of incidence: θ 1’ = θ 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.

Dr. G. Mirjalili, physics Dept. Yazd University

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. α =2β

α

β

Dr. G. Mirjalili, physics Dept. Yazd University

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.

Dr. G. Mirjalili, physics Dept. Yazd University

Refraction

All rays and the normal lie in the same plane.

• 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. • sinθ 2 / sinθ 1 = v2 / v1 θ 1 and θ 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. Dr. G. Mirjalili, physics Dept. Yazd University

Reflection by plane surfaces y

r1 = (x,y,z)

z r2= (-x,y,z) r1 = (x,y,z)

x r3=(-x,-y,z)

x r2 = (x,-y,z) Law of Reflection r1 = (x,y,z) → r2 = (x,-y,z) Reflecting through (x,z) plane Dr. G. Mirjalili, physics Dept. Yazd University

y r4=(-x-y,-z)

Refraction by plane interface & Total internal reflection n2 θ2

θ2

n1 > n2 θ1

θ1 θ C

θ1

θ1

P Snell’s law Dr. G. Mirjalili, physics Dept. Yazd University n1sinθ1=n2sinθ2

n1

Examples of prisms and total internal reflection 45o

45o

45o

Totally reflecting prism 45o

Porro Prism Dr. G. Mirjalili, physics Dept. Yazd University

Total internal reflection  Total internal reflection

n2 sin θ 2 , sin θ 2 = 1 when n2 / n1 > 1 & n2 = n1 sin θ1. Since sin θ1 = n1 When this happens, θ 2 is 90o and θ1 is called critical angle. Furthermore when θ1 > θ crit , all the light is reflected (total internal reflection).

Dr. G. Mirjalili, physics Dept. Yazd University

Total internal reflection  Optical fibers

Dr. G. Mirjalili, physics Dept. Yazd University

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: n1sinθ 1=n2sinθ 2 As light travels from one medium to another, its frequency does not change but its wavelength does. λ 1n1 = λ 2n2, or λ 1/λ 2 = v1/v2. Light slows on entering a medium – Huygens Also, if n → ∞ ν = 0 i.e. light stops in its track !!!!! Dr. G. Mirjalili, physics Dept. Yazd University

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% n1 = 1.5 n2 = 1.0

incoming ray (100%) 96%

image looks displaced due to jog

8% reflected in two reflections (front & back) 4% 92% transmitted Dr. G. Mirjalili, physics Dept. Yazd0.16% University 4%

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

Dr. G. Mirjalili, physics Dept. Yazd University

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

Dr. G. Mirjalili, physics Dept. Yazd University

Example: depth of a swimming pool

Pool depth s = 2m person looks straight down. θ

θ

1

2

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 Dr. G. Mirjalili, physics Dept. Yazd University

na sin θ1 = sin θ 2

θ 2

θ 1

L

tan θ1 =

L s

tan θ 2 =

L L = s − ∆s s '

→ s tan θ1 = ( s − ∆s ) tan θ 2 for small angles: tan ->sin

s sin θ1 = ( s − ∆s ) sin θ 2 s sin θ1 = ( s − ∆s ) na sin θ1 ∆s = s

na − 1 1 = (2 m) = 50 cm. na 4

Dr. G. Mirjalili, physics Dept. Yazd University

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

L

n1 n2 n2 = − ⇒ s' = − s s s' n1 | s ' | tan θ1 = L, | s | tan θ 2 = L → s ' tan θ1 = s tan θ 2 tan θ ≈ sin θ ≈ θ for θ << 1

s’ s

⇒ s ' sin θ1 = s sin θ 2 ⇒ n1 s ' = n2 s ( n1 sin θ1 = n2 sin θ 2 )

Dr. G. Mirjalili, physics Dept. Yazd University

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) sin θ n1 = 1

1

n2 sin θ

60°

= n2 sin θ

2

= sin θ

4

3

Example: θ 1 = 30° θ1

α

θ2

β θ3

 sin(30)  o θ 2 = sin −1   = 19.5  1.5  θ4

θ 3 = (60 o − θ 2 ) = 40.5o θ 4 = sin −1 (1.5 sin θ 3 ) = 76.9 o

n2 = 1.5

θ

3

= 90° - β

α +β +60o = o → α = 90° - θ 180 θ 2

3

= 60° - θ

Dr. G. Mirjalili, physics Dept. Yazd University

2

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” Dr. G. Mirjalili, physics Dept. Yazd University

Dispersion &Deviation n1
n1

n2

Little dispersion

High dispersion

High deviation

Low deviation

Dr. G. Mirjalili, physics Dept. Yazd University

Angular Dispersion A hollow 600 prism is filled with

δ

carbon disulfide, whose index of refraction for blue is 1.652, for red light is 1.618 what is angular 1 dispersion sin (σ + δ ) n prism n0

n1=1.652 δ 1=51.380

=

n2=1.618 δ 2=48.00

δ 1-δ 2 =3.380 angular dispersion Dr. G. Mirjalili, physics Dept. Yazd University

2

1 sin α 2

Deviation & wavenumber in prism • Deviation angle & λ b

A

c B b+nt1+d=c+nt2+e

t1

d

A` A``

a t2

a∆ D

e

dD/dn = (t2-t1)/a

b+(n+ ∆ n)t1+d+a ∆ D=c+ (n+ ∆ n)t2+e dD/dn =t/a ∆ nt1+a ∆ D=∆ n t2 ∆ D/ ∆ =(t2-t1)/a

dD/dλ =(dD/dn)(dn/dλ )=(t/a) (dn/dλ ) (dn/dλ Dr. G. Mirjalili, physics Dept. Yazd)=? University

Deviation angle & λ n = A+B/λ 2 dn/dλ = -2B/λ 3 dD/dλ = t/a dn/dλ dD/dλ = t/a(-2B/λ 3)

Dr. G. Mirjalili, physics Dept. Yazd University

Dispersive power & Abbe`s nunber

n f − nc nD − 1

Dispersive power

nD − 1 = Abbe`s − number = ν n f − nC

Low dispersion, low refractive index

ν Dr. G. Mirjalili, physics Dept. Yazd University

Refractive indices of Crown and flint glasses Fraunhofer line

color

λ (nm) n crown n flint

F D C

Blue Yellow Red

486.1 589.3 656.3

1.5293 1.5230 1.55204

ν

Crown

=59 Dr. G. Mirjalili, physics Dept. Yazd University

1.7378 1.7200 1.7130

ν

flint=29

Dispersing prisms • Achromatic prism: • Deviates light but gives no dispersion

λ λ

Dispersion for λ zero

1

1

2

and λ

2

is

Dr. G. Mirjalili, physics Dept. Yazd University

Dispersing prisms • Direct-vision prism λ

1

λ λ 2

Direct vision for wavelength λ

Dr. G. Mirjalili, physics Dept. Yazd University

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) δ F = σ 1(nf - 1) δ C = σ 1(nc- 1) → δ f - δ c = σ 1 (nf - nc) mean dispersion of prism For the combination to be achromatic, (σ 1)(n1f - n1c ) + (σ 2)(n2f - n2c )=0 (14)(1.5293-1.5204)+ (σ 2)(1.7378-1.7130)=0 σ 2= -50

Dr. G. Mirjalili, physics Dept. Yazd University

Example (cont.) • (b) for the direct-vision prism δ 1=σ 1(n1D-1) δ 2=σ 2(n2D-1) σ 1(n1D-1)= σ 2(n2D-1) 14(1.5230-1)= σ 2(1.7200-1) σ 2=10.20

Dr. G. Mirjalili, physics Dept. Yazd University

Image manipulation by reflection prisms Right angle prism

Dr. G. Mirjalili, physics Dept. Yazd University

Image manipulation by reflection prisms Dove prism

Dr. G. Mirjalili, physics Dept. Yazd University

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.

Dr. G. Mirjalili, physics Dept. Yazd University

Diversion & dispersion  Examples

Dr. G. Mirjalili, physics Dept. Yazd University

Resolving power of a prism

T

F d

d n b

FT+TW=nb FT+ Tw - ∆ s= (n- ∆ n) b ∆ s=b ∆ n ∆ s=b (dn/dλ ) ∆ λ

∆ s W

∆α

∆ α

λ +∆ λ ∆ α = λ /d

λ

λ /d = (b/d)(dn/dλ )∆ λ (∆ λ )min = λ /b(dn/dλ ) R=λ /(∆ λ )min = b(dn/dλ )

Dr. G. Mirjalili, physics Dept. Yazd University ∆ α =∆ s/d=(b/d) (dn/dλ )∆ λ

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 λ =550 nm.

•

solution

∆ n/∆ λ =(nf-nD)/(λ f-λ D)= (1.7328-1.7205)/(486-587)=-1.9x10 -4 nm -1 R = b(dn/dλ ) = (0.05x10 9nm)(-1.9x10 -4 nm -1) = 5971 (∆ λ )min=λ /R =5500A0/5971 ≈ 1 A0

Dr. G. Mirjalili, physics Dept. Yazd University

• Exercises

Dr. G. Mirjalili, physics Dept. Yazd University

Example

Exercises

The prism shown in the figure has a refractive index of 1.66, and the angles A are 25.00 . Two light rays m and n are parallel as they enter m the prism. What is the angle between them they emerge? n

A A

Solution

na sin θ a −1 1.66 sin 25.0° na sin θ a = nb sin θ b → θ b = sin ( ) = sin ( ) = 44.6°. nb 1.00 Therefore the angle below the horizon is θb − 25.0° = 44.6° − 25.0° = 19.6°, and thus the angle between the two emerging beams is 39.2°. −1

Dr. G. Mirjalili, physics Dept. Yazd University

Exercises

Example

Light is incident in air at an angle on the upper surface of a transparent plate, the surfaces of the plate being plane and parallel to each other. (a) t Prove that θ a = θ a' . (b) Show that this is true for any number of different parallel plates. (c) Prove that the lateral displacement D of the emergent beam is given by the sin(θ a − θ b' ) relation: d =t ,

θa

n

θ b'

Q

n’ n

θb P θ a'

d

cos θ b'

where t is the thickness of the plate. (d) A ray of light is incident at an angle of 66.00 on one surface of a glass plate 2.40 cm thick with an index of refraction 1.80. The medium on either side of the plate is air. Find the lateral Displacement between the incident and emergent rays.

Dr. G. Mirjalili, physics Dept. Yazd University

Exercises

Problem Solution

(a) For light in air incident on a parallel-faced plate, Snell’s law yields: n sin θ a = n' sin θb' = n' sin θb = n sin θ a' → sin θ a = sin θ a' → θ a = θ a' .

θa

n t

θ b'

n’

Q

(b) Adding more plates just adds extra steps in the middle of the above equation that P n θb always cancel out. The requirement of L d θ a' θ n = θ n' 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)

t t sin(θ a − θ b' ) d = L sin(θ a − θ ), L = →d = . cos θ b' cos θ b' ' b

n sin θ a sin 66.0° ) = sin −1 ( ) = 30.5° n' 1.80 ( 2.40cm ) sin(66.0° − 30.5°) →d = = 1.62 cm. cos 30.5°

θ b' = sin −1 (

Dr. G. Mirjalili, physics Dept. Yazd University