南开大学光学工程内部课件Oct 19th

The reflected ray can be treated as a ray from the image S’ below the mirror
Interference Pattern from Lloyd’s Mirror
In order to get stable intensity-modulated interference pattern by using two beams, two conditions must be satisfied
The two waves must maintain a constant phase with respect to each other, i.e. have the same frequency and constant initial phase The two electric fields have the non-zero projective components onto each other, i.e. the polarization directions cannot be perpendicular to each other
Interference Equations

Fra Baidu bibliotek
When total destructive interference occurs, a dark fringe is observed
This needs a path difference of an odd half wavelength δ = d sin θdark = (m + ½ ) λ


This is destructive interference

A dark fringe occurs
Young’s Experiment (Double-Slit Interference)
Interference Equations

The path difference, δ, is found from the tan triangle Optical Path Differenceδ = L = r2 – r1 = d sin
Iinter * E1 E2 c.c.
I1 E
2 1
,
I2 E
2 2
Interference Pattern
I I1 I 2 2 I1I 2 cos cos
where presents the crossing angle between the linear polarization directions of the two beams， and
m


= 0, ±1, ±2, …
Interference Equations

Y:measured vertically from the zeroth order maximum Assumptions


L >>d,
d >>λ

y =LtanθLsinθ
I I1 I 2 2 I1I 2 cos(kd sin ) I1 I 2 2 I1I 2 cos(kdy / L)
d 7 /(n 1) 6.6 m
Lloyd’s Mirror

Produce an interference pattern with a single light source Wave reach point P either by a direct path or by reflection
Two beam interference
Superposition of two beams
P
Assume the observation point P is far enough away from the sources so that the wavefronts at P can be regarded as planes. And we limit our discussion to linearly polarized waves.

The transmitted light is incident onto a screen containing two narrow slits
Young’s Double Slit Experiment

The symmetric narrow slits, S1 and S2 act as the two light sources The waves from the two slits come from the same source S0 and therefore are always in phase.
E1 (r , t ) E01 exp i(k1 r t 1 ) E2 (r , t ) E02 exp i(k2 r t 2 ).
Two plane waves meet at P
Superposition of two beams

Other Coherent Sources

Currently, it is much more common to use a laser as a coherent source The laser produces an intense, coherent,

monochromatic parallel beam over a width of several millimeters
mica, thickness d
1
Phase with meca
1 k ' d 2 nd /
Phase without meca
d
2
2 kd 2 d /
air, thickness d
2 d (n 1) / 7 (2 )
The mica thickness is :
P
Obviously,
2 * * * I E E E ( E1 E2 ) ( E1 E2 ) * * * * E1 E1 E2 E2 E1 E2 E2 E1 2 2 * E1 E2 E1 E2 c.c. I1 I 2 I inter
Interference Equations

For bright fringes
ybright

L
d
m m 0, 1, 2
For dark fringes
ydark
1 m m 0, 1, 2 d 2
L
Usage of Young’s Double Slit Experiment

Light from a monochromatic source goes through a narrow slit. The narrow width of the silt make sure transmitted light comes from a tiny region of the source which is coherent.
Laser pointer
Young’s double-slit interference
Side view of the interference field
Interference Patterns

Constructive interference occurs at the center point The two waves travel the same distance

Young’s Double Slit Experiment

In 1801, Thomas Young demonstrated for the first time interference of light The coherent sources was gotten through:


Young’s Double Slit Experiment provides a method for measuring wavelength of the light
This experiment gave the wave model of light a great deal of credibility
Interference Equations
This assumes the paths are parallel Not exactly, but a very good approximation (L>>d)

Interference Equations
I I1 I 2 2 I1I 2 cos(k ) I1 I 2 2 I1I 2 cos(kd sin )
It

is inconceivable that particles of light
could cancel each other
A thin flake of mica (n = 1.58) is used to cover one slit of a double-slit arrangement. The central point on the screen is then occupied by what used to be the seventh bright fringe. If = 550 nm, what is the thickness of the mica? The 7th bright fringe with no mica corresponds to n = 7 The mica therefore introduce a path difference of 7 or a phase difference of 7x2 at the screen center


Therefore, the waves arrive in phase

A bright fringe occurs
Interference Patterns

The upper wave travels one-half of a wavelength farther than the lower wave The trough of the bottom wave overlaps the crest of the upper wave (180 phase shift)


Therefore, they arrive in phase
Interference Patterns

The upper wave has to travel farther than the lower wave The upper wave travels one wavelength farther

For a bright fringe interference happens
where
total
constructive
2m k m

m = 0, ±1, ±2, … m is called the order number
When When
m = 0, it is the zeroth order maximum m = ±1, it is called the first order maximum
k1 r 1 k2 r 2 .
parallel
I I1 I 2 2 I1I 2 cos
Relation between Polarization Directions
perpendicular
I I1 I 2
Coherence Conditions for Interference