Aoptics-1-wave Fronts

Geometrical Optics p

Applied optics Wave fronts

Dr. G. Mirjalili, physics Dept. Yazd University

W Wavefronts f t • 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: cc. • Wavefronts propagate perpendicular to the local wavefront surface. Dr. G. Mirjalili, physics Dept. Yazd University

In describing the propagation of light g as a wave we need to understand: wavefronts: a surface passing th through h points i t off a wave that th t have the same phase and p amplitude. rays: a ray describes the direction of wave propagation. A ray iis a vector t perpendicular di l to the wavefront.

Dr. G. Mirjalili, physics Dept. Yazd University

Light Rays • The propagation of the wavefronts can be described by light rays rays. • In free space, the light rays travel in straight lines 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 p p g of light. g • 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 rays.

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 Pl

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

Huygens’ principle (cont’d) ‰ Huygens’ principle for plane wave • • •

At t = 0, 0 the th wave ffrontt is i indicated by the plane AA’ The points are representative sources for f the th wavelets l t 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) • •

•

Th inner The i arc represents part of the spherical wave The p points are representative p 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 Th l off reflection fl ti can be b derived from Huygen’s Principle AA’ iis a wave ffrontt off incident i id t light The reflected wave front is CD

Triangle ADC is congruent to triangle AA’C AA C Angles θ1 = θ1’ This is the law of reflection Dr. G. Mirjalili, physics Dept. Yazd University

Reflection

Huygens’ principle (cont’d) ‰ Huygens’ principle for law of refraction • •

IIn time ti ∆t ray 1 moves from ∆t, f A to B and ray 2 moves from A’ to C F From triangles ti l AA’C and d ACB ACB, all the ratios in the law of refraction can be found: n1 sin θ1 = n2 sin θ2

l sin θ1 = v1∆t; l 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

Reflection and refraction ‰ Reflection (cont’d)

• 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

l = AC

Dr. G. Mirjalili, physics Dept. Yazd University

The Law of reflection • Law of reflection: The angle of reflection equals the angle of incidence: θ1’ = θ1. • Some definitions: – Normal: The normal is a line drawn perpendicular to the surface at the point where the incident ray strikes strikes. – Angle of reflection and incidence: Measured from the normal to the reflected and incident rays rays, respectively.

Example : The double-reflected light ray •

Two mirrors make an angle of 120° with each other 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

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

Practical applications of reflection •

Refraction •

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

•

• All rays and the normal lie in the same plane. Dr. G. Mirjalili, physics Dept. Yazd University

β

Refraction: When a ray of light traveling through a transparent medium encounters a boundary g into another transparent p medium,, p part of leading the ray enters the second medium. The part that enters the second medium is bent at the boundaryy and is said to be refracted. sinθ2 / sinθ1 = v2 / v1 – θ1 and θ2 are the angle of incidence and angle of refraction, refraction respectively respectively. – v1 and v2 are the speed of the light in the first and second medium, respectively. The path of a light ray ra through thro gh a refracting surface is reversible.

Dr. G. Mirjalili, physics Dept. Yazd University

Refraction by plane interface & Total internal reflection

Reflection by plane surfaces y

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

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

θ2

x y

r3=(-x,-y,z)

r4=(-x-y,-z)

x

n2 θ2

n1 > n2

r2 = (x,-y,z)

θ1

Law of Reflection

θ1 θ C

θ1

θ1

n1

r1 = (x,y,z) → r2 = (x,-y,z) P

Reflecting through (x (x,z) z) plane

Snell’s law n1sinθ1=n2sinθ2

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

Examples of prisms and total internal reflection

Total internal reflection ‰ Total internal reflection

45o

45o

45o

n2 sin θ 2 , sin θ 2 = 1 when n2 / n1 > 1 & n2 = n1 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) reflection). Since sin θ1 =

Totally reflecting prism 45o

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

Dr. G. Mirjalili, physics Dept. Yazd University

Total internal reflection

Index of Refraction and Snell’s Snell s Law of Refraction

‰ Optical fibers

•

• •

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

Atmospheric Refraction and Sunsets •

Let’s L ’ consider id a thick hi k piece i off glass l (n = 1.5), 1 5) and d the h lilight h paths h associated with it – reflection fraction = [( [(n1 – n2))/(n ( 1 + 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%

Light Li ht slows l on entering t i a medium di – Huygens H Also, if n → ∞ ν = 0 i.e. light stops in its track !!!!! Dr. G. Mirjalili, physics Dept. Yazd University

R fl ti Reflections, R Refractive f ti offset ff t •

Index of refraction n of a medium: n ≡ c/v – c = 3 x 108 m/s: speed of light in vacuum. vacuum – v: speed of light in the medium; v < c. – n > 1 for any medium and n = 1 for vacuum (or approximately in air) air). Snell’s law of refraction: n1sinθ1=n2sinθ2 As light travels from one medium to another, its frequency does not change but its wavelength does. – λ1n1 = λ2n2, or λ1/λ2 = v1/v2.

image g looks displaced p due to jog

8% reflected in two reflections (front & back)

•

Light rays from the sun are b bent as they h pass iinto the h atmosphere It is a g 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

4% 92% transmitted t itt d Dr. G. Mirjalili, physics Dept. Yazd0.16% University 4%

Dr. G. Mirjalili, physics Dept. Yazd University

Example: depth of a swimming pool

Mirages g

Pool depth s = 2m

•

• •

person looks straight down.

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 Th observer The b sees both b th an upright and an inverted image

the depth is judged by the apparent size of some object j of length g L at the bottom of the pool (tiles etc.)

θ2

θ1

s`is reduced distance

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

Dr. G. Mirjalili, physics Dept. Yazd University

na sin θ1 = sin θ 2 L tan θ1 = s tan θ 2 =

θ2

L L = s − ∆s s '

→ s tan θ1 = ( s − ∆s ) tan θ 2 f smallll angles: for l ttan ->sin > i

θ1

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

L

∆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 Th image i fforms between b t the object and the surface – The rays bend away from the normal since n1 > n2

L

n1 n n = − 2 ⇒ s' = − 2 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 (Q n1 sin θ1 = n2 sin θ 2 )

Dr. G. Mirjalili, physics Dept. Yazd University

Prism example

Prisms

• Light is refracted twice – once entering and once leaving. • Since n decreases for increasing λ, a spectrum emerges...

‰ Applications of prism

Analysis: l (60° glass prism in air) sin θ1 = n2 sin θ2 n2 sin θ3 = sin θ4

n1 = 1

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

60° Example: θ1 = 30°

α

θ1

θ2

β θ3

⎛ sin(30) ⎞ o ⎟ = 19.5 ⎝ 1.5 ⎠

θ 2 = sin −1 ⎜ θ4

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

n2 = 1.5 15

α+β+60o = 180o

θ3 = 90 90° - β

α = 90 90° - θ2

→

θ3 = 60 60° - θ2

• 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

Dr. G. Mirjalili, physics Dept. Yazd University

Dispersion &Deviation

Angular Dispersion

n1
A hollow 600 prism is filled with carbon disulfide, whose index of refraction for blue is 1.652, 1 652 for red light is 1.618 what is angular 1 dispersion sin (σ + δ ) n prism

n2

n0

n1=1.652

=

n2=1.618

Little dispersion

High dispersion

δ1=51.380

High deviation

Low

δ1-δδ2 =3.38 3 380 a angular gu a dispersion d spe s o

δ2=48.00

deviation

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

2 1 sin α 2

Deviation angle & λ

Deviation & wavenumber in prism • Deviation angle & λ b

A

c

t1

a∆D

d

A` A``

a t2

B

n = A+B/λ2 dn/dλ = -2B/λ3 dD/dλ = t/a dn/dλ dD/dλ = t/a(-2B/λ3)

e

b+nt1+d=c+nt2+e

dD/dn = (t2-t1)/a

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

dD/dn =t/a

∆nt1+a ∆D=∆n t2

dD/dλ=(dD/dn)(dn/dλ)=(t/a) (dn/dλ) (dn/dλ)=? (dn/dλ) ?

∆D/ ∆=(t2-tt1)/a

n f − nc nD − 1

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

Dispersive power & Abbe`s Abbe s nunber

Refractive indices of Crown and flint glasses

Dispersive power

nD − 1 = Abbe`s − number = ν n f − nC

Low dispersion dispersion, low refractive index

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

Fraunhofer line

color l

λ( ) λ(nm)

n crown n flint fli t

F D C

Blue Yellow Red

486.1 486 1 589.3 656 3 656.3

1.5293 1 5293 1.5230 1 55204 1.55204

1.7378 1 7378 1.7200 1 7130 1.7130

ν Crown =59 ν flint=29 Dr. G. Mirjalili, physics Dept. Yazd University

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

Dispersing prisms • Direct-vision prism λ1 λ λ2

λ1 λ2

Direct vision for wavelength g λ

Dispersion for λ1 and λ2 is zero Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

example

Example (cont (cont.))

• Assume that 140 is the apex angle of a crown glass prism. i Wh Whatt should h ld b be th the apex angle l off a fli flintt prism: i (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 = σ1(nf - 1) → δ f - δc = σ1 ((nf - nc) mean dispersion p δ C = σ1((nc- 1)) of prism For the combination to be achromatic, (σ1)(n1f - n1c) + (σ2)(n2f - n2c)=0 (14)(1.5293-1.5204)+ (σ2)(1.7378-1.7130)=0 σ2= -5 50 Dr. G. Mirjalili, physics Dept. Yazd University

• (b) for the direct-vision prism δ1=σ σ1((n1D-1)) δ2=σ2(n2D-1) σ1(n1D-1)= σ2(n2D-1) 14(1 5230-1)= 14(1.5230 1) σ2(1.7200 (1 7200-1) 1) 0 σ2=10.2

Dr. G. Mirjalili, physics Dept. Yazd University

Image manipulation by reflection prisms

Image manipulation by reflection prisms Dove prism

Right angle prism

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

Dispersion ‰ Dispersion

Diversion & dispersion ‰ Examples

• The index of refraction of a material depends on wavelength as shown on the right right. This is called dispersion dispersion. • It is also true that, although the speed off light li ht iin vacuum d does nott d depends d on wavelength, in a material, wave speed depends on wavelength.

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

Resolving power of a prism

T

F

d

d

n

∆s W ∆α ∆ ∆α λ+∆λ

b FT+TW=nb FT+ Tw - ∆s= (n- ∆n) b ∆s=b ∆n

∆α = λ/d

λ

λ/d = (b/d)(dn/dλ)∆λ

∆s=b (dn/dλ) ∆λ

(∆λ)min= λ/b(dn/dλ)

∆α=∆s/d=(b/d)(dn/dλ)∆λ

R=λ/(∆λ)min= b(dn/dλ)

Resolving power of a prism (example) •

A prism i made d ffrom flint fli glass l with ihab base off 5 cm. find fi d the h resolving l i power of the prism at λ=550 nm.

•

solution

∆n/∆λ=(nf-nD)/(λf-λD)= (1.7328-1.7205)/(486-587)=-1.9x10 -4 nm -1 R = b(dn/dλ) = (0.05x10 (0 05x10 9nm)(-1.9x10 nm)( 1 9x10 -44 nm -11) = 5971 ((∆λ))min=λ/R =5500A0/5971 ≈ 1 A0

Dr. G. Mirjalili, physics Dept. Yazd University

Dr. G. Mirjalili, physics Dept. Yazd University

Example

•Exercises

Exercises

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

A A

Solution

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

Dr. G. Mirjalili, physics Dept. Yazd University

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 plate, plane and parallel to each other. (a) t Prove that θ a = θ a' . (b) Show that this i ttrue ffor any number is b off different diff t parallel ll l plates. (c) Prove that the lateral displacement D of the emergent beam is given by the sin(θ a − θ b' ) relation: d =t ,

Problem θa

n

n

Solution (a) For light in air incident on a parallel-faced parallel faced plate, Snell’s law yields:

θb'

Q

n’

n sin θ a = n' sin θ b' = n' sin θ b = n sin θ a' → sin θ a = sin θ a' → θ a = θ a' .

θb P θ a'

d

cos θ b'

where t is the thickness of the p plate. ((d)) A ray y of light g is incident at an angle g 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 rays.

Dr. G. Mirjalili, physics Dept. Yazd University

Exercises θa

n t

n’

θb'

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' parallel ll l faces f ensures that th t th the angle l θn = θ n' and the chain of equations can continue. (c) The lateral displacement of the beam can be calculated using geometry: d = L sin(θ a − θ b' ), L =

(d)

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

sin 66.0° n sin θ a ) = sin −1 ( ) = 30.5° 1.80 n' ( 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