Z = p / u = ρ c

, with pressure amplitude

p

u

. Continuity of pressure and normal velocity at normal incidence on a lossless interface gives

r_{p} = \frac{Z _{2} - Z _{1}}{Z _{2} + Z _{1}}, R = r_{p}^{2}, T = \frac{4 Z _{1} Z _{2}}{( Z _{1} + Z _{2} ) ^{2}} .

(1)

Matching

Z_{1} = Z_{2}

eliminates reflection. A quarter-wave intermediate layer with

Z_{m} = Z_{1} Z_{2}

transforms the load to the source impedance at its design frequency; its bandwidth is limited. For strings, the analogous mechanical impedance is transverse force divided by velocity,

Z = T v^{- 1} = T μ

; continuity of displacement and transverse force gives the corresponding matching condition. These amplitudes use different force/velocity conventions from acoustic pressure, so reflection signs must be stated consistently.

b

 0  0

Solution

 0  0

Reflections from the two film surfaces interfere with propagation phase

4 πn d cos θ_{t} / λ_{0}

plus the relative reflection phase. Reflection from a lower to a higher refractive index contributes a

π

phase reversal. For a higher-index film surrounded by air there is one relative reversal, hence

reflection maxima: 2 n d cos θ_{t} = (m + \frac{1}{2}) λ_{0}, minima: 2 n d cos θ_{t} = m λ_{0} .

(1)

Thickness or angle variation therefore produces fringes or colours under broadband illumination. An antireflection coating between indices

n_{0}, n_{s}

instead uses, at normal incidence,

n_{f} = n_{0} n_{s}

and

n_{f} d = λ_{0} /4

: the two reflected amplitudes then cancel at the design wavelength. Absorption, polarization, unequal amplitudes and multiple reflections determine the full contrast.

c

 0  0

Solution

 0  0

A Michelson interferometer divides light at a beam splitter, sends it to two mirrors, and recombines the returned beams. A compensating plate can equalize glass traversal. For approximately axial rays the optical path difference is