Chap. 4 Multiple-Beam Interference

4.1 Interference with Multiple Beam

- Consider two semi-reflecting mirrors that are considered to be thin and identical.

- The geometrical path difference any two successive transmitted rays is cos.

The corresponding phase difference cos

cos

cos

- Then,

⋅⋅⋅ →

- The intensity of the transmitted light is

In general, r is a complex number;

where is the phase change for one reflection.

- The phase change for a dielectric is either 0 or , depending on the relative index of refraction (see Sec. 2.7); but in the case of a metal film, the phase change can be any value (see Sec. 6.5).

Let , then

with

cos

sin

- The formula for the intensity

sin

;

Airy function sin

and the coefficient of finesse

- sinhas the maxima at .

- As → (small), F is small and the interference fringes are broad and indistinct.

- The condition for a fringe maximum is equivalent to

coswith N = the order of interference.

- If two reflecting surfaces are NOT identical,

,

- For non-zero absorption at each reflection (A), we must have

The maximum intensity max

4.2 The Fabry-Perot Interferometer

- 1899, C. Fabry and A. Perot: to measure wavelengths with high precision.

- Two optically flat, partially reflecting plates of glass or quartz with their reflecting surfaces held accurately parallel (see Fig. 4.4). The plate spacing can be mechanically varied (interferometer) or held fixed by spacers (etalon). A flatness of the order of 1/20 to 1/100 wavelength is required!

- Free spectral range: →

cos →

for small

4.3 Resolution of Fabry-Perot Instruments

- Suppose a spectrum consisting of two closely spaced frequencies and ′ to be analyzed.

- Assuming that two components are of equal intensities. the fringe pattern is given by the sum of two Airy functions,

sin

sin

′

with

and

≈

, ′≈ ′

′ for small .

- If there is a dip in the intensity curve, the two frequencies ( ′) can be resolved.

☛ Taylor Criterion: two equal lines are considered to be resolved if the individual curves

cross at the half-intensity point, so that the total intensity at the saddle point is equal to the maximum intensity of each line alone.

sin

′

→ sin

′

For small ′, ′

In terms of the angular frequency, ′

Resolving power (RP) =

; RP =

4.4 Theory of Multilayer Films

- Optical surfaces having any desired reflectance and transmission characteristics; optical filters, anti-reflecting coatings, etc.

- Boundary conditions: the electric and magnetic fields should be continuous at each interface. For normal incidence, (1) First interface; Electric field ′ ′ Magnetic field ′ ′ or ′ ′ (2) Second interface; Electric field

′

Magnetic field ′

or ′

- Eliminating ′,

′ cos

sin

′ sin cos

or, in matrix form,

′

cos

sin

sin cos

- Simplified form

with reflection

′ and transmission

transfer matrix

cos

sin

sin cos

where

- Then, for N layers having indices of refraction and thicknesses ,

Let

, then solve the above equation for r and t;

→ and

→

● Antireflecting Films: for a single film of index and thickness l, placed on a glass substrate of index

With , cos

sin

cos sin

- If the optical thickness of the film is 1/4 wavelength, then ;

→ if

With magnesium fluoride (MgF) , glass ≈ , for a quarter-wave film of MgF on a glass, about 1 % of reflectance (1/4 of the uncoated glass)

- Use of two layers, one of high index and the other of low index; nearly zero reflectance available for given wavelength → more layers for zero reflectance at broader wavelengths

● High-Reflectance Films: a stack of alternating layers of high index () and low index () with the thickness of each layer being 1/4 wavelength

- For two adjacent layers, transfer matrix is

- For the stack consisting of 2N layers,

Assuming that and , the reflectance of a multilayer stack is given by

[Homework]

Prob. #4.4, #4.9