Lecture 26-1 Lens Equation ( < 0 ) True for thin lens and paraxial rays. magnification m = h’/h = - q/p.

Lecture 26-Lecture 26-11 Lens Equation

1 1 1

p q f

( < 0 )

True for thin lens and paraxial rays.

magnification m = h’/h = - q/p

Lecture 26-Lecture 26-22Signs in the Lens Equation for Thin Lenses

1 1 1

p q f q

mp

• p is positive for real object

• q is positive for real image

• q is negative for virtual image

• m is positive if image is upright

• m is negative if image is inverted

• f is positive if converging lens

• f is negative if diverging lens

i

o

y f q q

y f p

• p is negative for virtual object

Lecture 26-Lecture 26-33Aberrations

Chromatic aberration

correct

Spherical aberration

nblue > nred

Parabolic mirrror

Cameras, …

Large telescopes, …

Lecture 26-Lecture 26-44Geometric Optics vs Wave Optics

• Geometric optics is a limit of the general optics where wave effects such as interference and diffraction are negligible.

Geometric optics applies when objects and apertures involved are much larger than the wavelength of light.

In geometric optics, the propagation of light can be analyzed using rays alone.

• Wave optics (sometimes also called physical optics) - wave effects play important roles.

Wave optics applies when objects and apertures are comparable to or smaller than the wavelength of light.

In wave optics, we must use the concepts relevant to waves such as phases, coherence, and interference.

Lecture 26-Lecture 26-55 Coherence When the difference in phase between two (or more) waves remains constant (i.e., time-independent), the waves are said to be perfectly coherent.

Only coherent waves can produce interference fringes!

Coherence length: the spatial extent over which light waves remain coherent.

A single light wave is said to be coherent if any two points along the propagation path maintains a constant phase difference.

- laser light and light transmitted through a small aperture are coherent.

- light from a light bulb and sun light over some area are incoherent.

Lecture 26-Lecture 26-66Interference of Two Coherent Waves

Constructive interference (in phase)

Snapshot of wave fronts at a given instant

Destructive interference (completely out of phase)

20

21 2( )

I c E

E E

B,C A

Lecture 26-Lecture 26-77Intensity of Interference Fringes

Let the electric field components of the two coherent electromagnetic waves be

1 0

2 0

sin

sin( )

E E t

E E t

The resulting electric field component point P is then

1 2

0

0

sin sin( )

2 cos sin2 2

E E E

E t t

E t

2

0 0

202

4 cos2

m II E

I EI

Intensity is proportional to E2

I=0 when = (2m+1) , i.e. half cycle + any number of cycle.

Lecture 26-Lecture 26-88Thin Film

• Thin here means that the thickness is comparable to the wavelength of the light.

• The reflected light waves from the two sides of a thin film interfere.

Phase change by

• If the incident light propagates from a medium of lower index of refraction toward one of higher index of refraction, the phase of the reflected wave shifts by .

(neither highlow index nor transmitted light)

1 2 3 2,n n n n

• Phase difference could come from: reflection, path length difference, different indices of refraction

Destructive interference eliminates (or minimizes) the reflected light!

e.g., non-reflecting lens coating

Lecture 26-Lecture 26-99Thin-Film Interference-Cont’d

Path length difference:

2l t

(Assume near-normal incidence.)

( 1/ 2)

m

m

destructive

constructive

0

n

where

• ray-one got a phase change of 180o due to reflection from air to glass.

• the phase difference due to path length is:

•then total phase difference:= ’+180.

22'0

n

ll

Lecture 26-Lecture 26-1010Thin-Film Continued

The previous discussion was for the situation in which n2 > n1 and n2 > n3 , i.e., the index of refraction of the film is larger than those of the surrounding media, but they are also valid if the index of refraction of the film is smaller than those of the surrounding media (n2 < n1 and n2 < n3 ). The equations fail for some of the following situations. Which one(s)?

If the film has an intermediate index of refraction

Conditions for maxima/minima will reverse!

Lecture 26-Lecture 26-1111

Warm-up quiz

a) The reflection is dark by destructive interference of rays 1 and 2

b) The reflection is bright by constructive interference of rays 1 and 2

c) The reflection is colorful by interference of rays 1 and 2.

2

1

water (n=1.33)

oil (n1>1.33)

Monochrome light of wavelength in air is incident normal to a thin layer of oil film floating on water as shown. If the film thickness is 5/(4n1). Which of the following statement is true?


The air between the glass plates acts like a thin film.

• Since the thickness of the film changes over the radius of the plates, alternating bright and dark fringes form, when the plates are illuminated. Because of the curvature of the upper piece, the film thickness varies more rapidly at larger radius. Thus the fringe separation is smaller toward the outside.

Newton’s Ring

Lecture 26-Lecture 26-1313 Two (narrow) slit Interference

• Upon reaching the screen C, thetwo wave interact to produce aninterference pattern consisting ofalternating bright and dark bands(or fringes), depending on theirphase difference.

Constructive vs. destructiveinterference

• According to Huygens’s principle,each slit acts like a wavelet. The the secondary wave fronts arecylindrical surfaces.

Young’s double-slit experiment

Lecture 26-Lecture 26-1414 Interference Fringes

For D >> d, the difference in path lengthsbetween the two waves is

sindL • A bright fringe is produced if the path lengths differ by an integer number of wavelengths,

sin , 0, 1,d m m

• A dark fringe is produced if the path lengths differ by an odd multiple of half a wavelength,

sin ( 1/ 2) , 0, 1,d m m

y ~ D*tan(θ)

Lecture 26-Lecture 26-1515Intensity of Interference Fringes

Let the electric field components of the two coherent electromagnetic waves be

1 0

2 0

sin

sin( )

E E t

E E t

The resulting electric field component point P is then

1 2

0

0

sin sin( )

2 cos sin2 2

E E E

E t t

E t

2

0 0

202

4 cos2

m II E

I EI

Intensity is proportional to E2

I=0 when = (2m+1) , i.e. half cycle + any number of cycle.


Intensity of Interference Fringes-Cont’d

For Young’s double-slit experiment, the phase difference is

2 , maxima2 ( 1/2), mimina

sin2 m for

m ford

204 cos

2I


Physics 241 –Quiz A

a) /(4n1)

b) /4

c) /(2n1)

d) /2

e) 3/4

2

1

glass (n1>1)

Light of wavelength in air is incident normal to a thin layer of glass held in air as shown. If the reflection is suppressed (dark) by interference of rays 1 and 2, what is a possible thickness d of the glass layer?


Physics 241 –Quiz B

a) /(4n1)

b) /4

c) /(2n1)

d) /2

e) 3/4

2

1

water (n=1.33)

oil (n1>1.33)

Light of wavelength in air is incident normal to a thin layer of oil film floating on water as shown. If the reflection is suppressed (dark) by interference of rays 1 and 2, what is a possible thickness d of the oil film?


Physics 241 –Quiz C

a) /(4n1)

b) /4

c) /(2n1)

d) /2

e) 3/4

2

1

water (n=1.33)

oil (n1>1.33)

Light of wavelength in air is incident normal to a thin layer of oil film floating on water as shown. If the reflection is bright (constructive interference) by interference of rays 1 and 2, what is a possible thickness d of the oil film?