Question

1. A Fabry-Perot etalon (or resonator) is formed by two mirrors in air. The distance between mirrors is 1.25 (mm), and the reflectance of each mirror is 0.92.

What is the mode of the etalon that is closest to the wavelength of 1.55 (µm)?

2. What is the separation of the modes in the etalon?

3. The finesse, *F*, of the etalon is defined as the
ratio of mode separation to spectral width of each mode, and
depends on the mirror reflectances, *R*, as:

F=π√R/(1-R)

Find the finesse of this etalon {give your answer to 4 significant digits, this is a unitless quantity}.

4. What is the spectral width of each mode, δVm (in GHz)

5. What is the quality factor for this resonator?

Answer #1

Assume an air-filled Fabry-Perot interferometer made from two
similar, parallel mirrors. For a wavelength of 610nm, the fraction
of transmitted light to incident light is 0.0154. The distance of
separation is ℓ = 2.60μm. What is (a) the finesse, F, and (b) the
reflectance of the mirrors, R, when assuming only reflective
contributions to the etalon's finesse?

A Fabry-perot interferometer has a 1.4cm spacing between the
mirrors and a reflection
coefficient of r = 0.95. For a wavelength around 532nm;
(a) (5 points) What is the minimum resolvable wavelength interval
△λ?
(b) (5 points) What is the resolving power R?
(c) (5 points) What is the coefficient of finesse F?
(d) (5 points) With this resolving power found in part (b), what is
the order of peak
m?

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